ELEMENTS 


ALGEBRA 


MAJOR    D.   H.   HILL, 

PEOFESSOK  OP  MATHEMATICS  AND  CIVIL  ENQINEEHnfO  IN  DAVIDSON  COLLEGE,  N.  C. 
LATE  FROFESSOa  OF  MATHEMATICS  IN  WASHINGTON  COLLBOE,  VA. 


PHILADELPHIA: 
J.   B.   LIPPINCOTT    &    CO. 

1857. 


Entered,  according  to  the  Act  of  Congress,  in  the  year  1857,  by 
J.  B.  LIPPINCOTT    &    CO., 
in  the  Clerk's   Office  of  the  District  Court  of  the   United   States  for  the  Eastern 
District  of  Pennsylvania. 

STEREOTYPED  BY  J.  FAGAN. 


TESTIMONIALS. 


From  T.  J.  Jackson,  Professor  of  Natural  and  Experimental  Philo- 
sophy, Virginia  Military  Institute. 

"  Froin  an  examination  of  various  portions  of  Major  D.  H.  Hill's  Algebra,  in 
manuscript,  I  regard  it  as  superior  to  any  other  work  with  which  I  am 
acquainted  on  the  same  branch  of  science." 


From  a  Teacher  of  Mathematics. 

"  Having  also  examined  several  chapters  of  Major  Hill's  Algebra,  I  have  no 
hesitation  in  concurring  in  the  above  opinion  of  Professor  Jackson." 

WILLIAM  Mclaughlin. 


From  J.  L.  Campbell,  Professor  of  Natural  Science,  Washington 
College,  Virginia. 

"  While  I  fully  concur  with  Professor  Jackson  and  Mr.  McLaughlin  in  the 
opinion  they  express  of  Major  Hill's  Algebra,  I  will  add,  that  I  regard  the 
method  adopted  by  the  author,  of  incorporating  into  the  work  some  of  the 
elementary  principles  of  the  Calculus,  as  giving  it  peculiar  value  as  a  college 
text-book." 


From  William  Gilham,  Professor  of  Chemistry  and  Geology,  Virginia 
Military  Institute,  and  late  Professor  at  West  Point,  iV.  Y. 

"  Having  read  the  greater  portion  of  Mnjor  Hill's  Algebra,  I  consider  it  as 
better  adapted  for  use  as  a  college  text-book  than  any  work  on  the  subject  with 
which  I  am  acquainted." 


From  Professor  J.  A.  Leland,  late  Professor  of  Mathematics  in  the 
State  Military  Institute  of  S.  C. 

"  I  have  examined  with  care,  many  of  the  proof  sheets  of  Major  Hill's  Algebra, 
and  it  affords  me  pleasure  to  concur  in  the  favorable  opinions  above  expressed." 


105971 


IV  TESTIMONIALS. 

This  work  of  Professor  Hill's  is  the  product  of  a  mind  intensely  in  love  with 
Algebra.  It  bears  the  marks  of  unremitting  and  intelligent  toil.  It  is  exhaus- 
tive on  the  subjects  it  treats ;  and,  in  the  abundance  and  aptness  of  its  illustra- 
tious,  reminds  one  of  the  richness  and  simphcity  of  Euler. 

CHARLES  PHILLIPS, 
Prof.  CM  Engineering,  University,  N.  C. 


CONTENTS 


TAQB 

DEFINITIONS  AND  PRELimNARY  REMARKS 13 

ADDITION 18 

Case  L    When  the  Terms  are  like,  and  have  like  Signs 18 

XL     When  the  Quantities  are  like,  but  aflFected  with  unlike  Signs..  19 
III.     When  the  Quantities  are  similar  and  dissimilar,  and  have  like 

and  unlike  Signs 20 

SUBTRACTION 21 

MULTIPLICATION 24 

Case  I.     When  both  Multiplicand  and  Multiplier  are  Simple  Quantities  24 
n.     When  the  Multiplicand  is  a  Compound  Quantity,  and   the 

Multiplier  a  Monomial 27 

III.     When  the  Multiplicand  and  Multiplier  are  both  Compound 

Quantities 29 

Theorems 33 

Factoring  Folynomials 35 

DIVISION '. 38 

Case  I.     When  the  Dividend  and  Divisor  are  both  Monomials 38 

II,     When  the  Dividend  is  a  Compound  Quantity,  and  the  Divisor 

a  Monomial 41 

III.     When  the  Dividend  and  Divisor  are  both  Polynomials 42 

Principles  in  Division 48 

ALGEBRAIC  FRACTIONS 51 

Reduction  of  Fractions 54 

Case  I.     To  reduce  a  Simple  Fraction  to  its  Simplest  Form 54 

II.  To  reduce  a  Mixed  Quantity  to  the  Form  of  a  Fraction 55 

III.  To  reduce  a  Fraction  to  an  Entire  or  j\Iixed  Quantity 56 

1*  (V) 


Yl  CONTENTS. 

_,    ,  ^  PAGE 

Reduction  of  Fractions  (continued).  ^ 

Ca^e  IV.     To  develop  a  Fraction  into  a  Series 57 

V.     To  reduce  Fractious  having  Diiferent  Denominators  to  Equi- 
valent Fractions  having  the  same  Denominator 69 

VI.     To  add  Fractional  Quantities  together 61 

VII.     To  Subtract  one  or  more  Fractional  Quantities  from  one  or 

more  Fractional  Quantities 62 

VIII.     To  multiply  Fractional  Quantities  together 64 

IX.     To  divide  Fractional  Quantities  by  each  other 65 

X.     To  reduce  a  Compound  Fraction  to  its  Lowest  Terms 68 

Least  Common  Multiple 76 

Corollary 78 

Least  Common  Multiple  of  Fractions 79 

Greatest  Common  Divisor  of  Fractions 80 

EQUATIONS  OF  THE  FIRST  DEGREE 82 

Solution  of  Equations  of  the  First  Degree 85 

First  Transformation  —  clearing  an  Equation  of  Fractions 85 

Second  Transformation — transposing  Terms 86 

Third  Transformation  —  clearing  the  Unknown  Quantity  of  Coefficients  86 
Solutions  of  Problems  producing  Equations  of  the  First  Degree  with  an 

Unknown  Quantity 91 

GEOMETRICAL  PROPORTION 95 

Theorems 96 

General  Remarks 103 

Problems  in  Geometrical  Proportion 105 

NEGATIVE  QUANTITIES 107 

Problem  of  the  Couriers ; 110 

General  Problems '. 116 

ELIMINATION    BETWEEN    SIMULTANEOUS   EQUATIONS    OF    THE 

FIRST  DEGREE 129 

Elimination  by  Comparison 130 

Elimination  by  Addition  and  Subtraction 131 

Elimination  by  the  Greatest  Common  Divisor 133 

Examples  in  Elimination  between  two  Simple  Equations  of  the  First 

Degree  involving  two  Unknown  Quantities 135 

Elimination  between  any  Number  of  Simultaneous  Equations 140 

General  Remarks 149 

Problems  producing  Simultaneous  Equations  of  the  First  Degree 148 


CONTENTS.  VU 

PAGE 

VANISHING  FRACTIONS 159 

FORMATION  OF  THE  POWERS  AND  EXTRACTION  OF  ROOTS 163 

Of  Incommensurable  Numbers 175 

Extraction  of  the  Square  Root  of  Fractions 177 

Mixed  Numbers 184 

Roots  of  Numbers  entirely  Decimal 186 

Square  Root  of  Fractions  expressed  decimnlly  187 

Extraction  of  the  Cube  Root  of  Numbers 188 

Approximate  Roots  of  Incommensurable  Numbers 19G 

Cube  Root  of  Fractions 197 

Approximate  Root  to  within  a  certain  Decimal 199 

Cote  I.     Approximate  Root  of  Whole  Numbers  to  within  a  certain 

Decimal 199 

II.     Approximate  Roots  of  Mixed  Numbers  to  within  a  cer- 
tain Decimal 200 

Approximate  Root  of  Decimal  Fractions  to  within  a  certain  Decimal....  202 

Approximate  Root  of  Vulgar  Fractions  to  within  a  certain  Decimal 203 

General  Remarks  on  the  Extraction  of  the  Cube  Root 204 

Extraction  of  the  Square  Root  of  Algebraic  Quantitiet 206 

Square  Root  of  Monomials 206 

Square  Root  of  Polynomials 207 

Square  Root  of  a  Polynomial  involving  Negative  Exponents 214 

Square  Root  of  Incommensurable  Polynomials 216 

Square  Root  of  Polynomials  containing  Terms  affected  with  Fractional 
Exponents 217 

Cube  Root  of  Polynomials 217 

Cube  Root  of  Incommensurable  Polynomials 222 

Cube  Root  of  Polynomials  involving  Fractional  Exponents 223 

Cube  Root  of  Polynomials  containing  one  or  more  Terms  affected  with 
Negative  Exponents 224 

Quantities  affected  with  Fractional  Exponents 225 

Multiplication  of  Quantities  affected  with  Entire  or  Fractional  Ex- 
ponents    227 

Division  of  Quantities  affected  with  Fractional  and  Entire  Exponents — 
Monomials 228 

Raising  to  Powers  Quantities  affected  with  Fractional  and  Entire  Ex- 
ponents—  Monomials 230 


VIU  CONTENTS. 

PAGE 

Quantities  affected  with  Fractional  Exponents  (continued). 
Extraction  of  Roots  of  Quantities  aifected  with  Entire  or  Fractional 

Exponents  —  Monomials 232 

Promiscuous  Examples 233 

Calculus  of  Radicals 236 

First  Principle 238 

Second  Principle 240 

Third  Principle 241 

Fourth  Principle 242 

Fifth  Principle 244 

Sixth  Principal 247 

SeTenth  Principle 248 

Eighth  Principle 250 

To  make  Surds  rational  by  Multiplication 253 

Case  I.     Monomial  Surds 253 

Corollary 254 

II.  To  find  a  Multiplier  that  will  make  rational  an  Expression 

consisting  of  a   Monomial  Surd  connected  with  Rational 

Terms,  or  consisting  of  two  Monomial  Surds 256 

Corollary 261 

III.  To  make  rational  an  Expression  containing  three  or  more 

Terms  of  the  Square  Root 262 

Corollary .". 264 

Extraction  of  the  Square  Root  of  a  Monomial  Surd  connected  with  a  Rational 

Term,  or  of  two  Monomial  Surds 265 

First  Principle 265 

Second  Principle 265 

Third  Principle 266 

Imaginary  Quantities 270 

EQUATIONS  OF  THE  SECOND  DEGREE 276 

Incomplete  Equations 276 

Properties  of  Incomplete  Equations 284 

Binomial  Equations 286 

General  Problems  in  Binomial  Equations 287 

Complete  Equations  of  the  Second  Degree 289 

First  Property.    Every  Complete  Equation  of  the  Second  Degree  has 
two  Values,  and  but  two,  for  the  Unknown  Quantity .., 295 


CONTENTS.  IX 

EQUATIONS  OF  THE  SECOND  DEGREE  — 

Complete  Equations  of  the  Second  Degi'ee  (continued). 

Second  Property.  The  First  Member  of  every  Equation  of  the  Second 
Degree  can  be  decomposed  into  two  Factors  of  the  First  Degree 
with  respect  to  x,  the  First  Factor  being  the  Algebraic  Sum  of  x 
and  the  First  Value  of  x  with  its  Sign  changed ;  and  the  Second 
Factor  being  the  Algebraic  Sum  of  x  and  the  Second  Value  with 
its  Sign  changed.  The  Second  Member  of  the  Equation  after  this 
Decomposition  will  be  Zero 296 

Third  Property.  The  Algebraic  Sum  of  the  Values  of  every  Com- 
plete Equation  of  the  Second  Degree  is  equal  to  the  Coefficient  of 
the  First  Power  of  the  Unknown  Quantity,  with  its  Sign  changed..  298 

Fourth  Property.  The  Product  of  the  Values  in  every  Complete 
Equation  of  the  Form,  x^  -\-  px  t=  q,  is  equal  to  the  Second  Mem- 
ber or  Absolute  Term,  with  its  Sign  changed 298 

Fifth  Property.  The  Value  of  x,  in  every  Complete  Equation  of  the 
Second  Degree,  is  half  the  Coefficient  of  the  First  Power  of 
X,  plus  or  minus  the  Square  Root  of  the  Square  of  half  this  Co- 
efficient increased  by  the  Absolute  Term 299 

General  Examples 300 

Discussion  of  Complete  Equations  of  the  Second  Degree 304 

Irrational,  Imaginary,  and  Equal  Values 307 

Suppositions  made  upon  the  Constants 308 

Explanation  of  Imaginary  Values 309 

Explanation  of  Negative  Solutions 311 

General  Problems  involving  Complete  and  Incomplete  Equations  of  the 

Second  Degree 311 

Trinomial  Equations 326 

Problem  of  the  Lights 329 

UNDETERMINED  COEFFICIENTS 335 

Failing  Cases 341 

Particular  Cases 345 

DERIVATION 347 

Theorems 351 


X  CONTENTS. 

PAGE 

BINOMIAL  FORMULA 356 

Demonstration 357 

Formation  of  Powers  by  the  Rule 360 

Demonstration  of  the  Binomial  Formula  for  any  Exponent 364 

Development  of  Binomials  aflFected  witli  Negative  and  Fractional  Ex- 
ponents   865 

Consequences  of  the  Binomial  Formula  —  Square  of  any  Polynomial...  367 

Cube  of  any  Polynomial 368 

Extraction  of  the  nth  Root  of  Whole  Numbers  and  Polynomials 369 

Approximate  Root  of  an  Irrational  Number  to  within  a  certain  Vulgar 

Fraction 372 

Approximate  Root  of  Whole  Numbere  to  within  a  certain  Decimal 373 

Approximate  wth  Root  of  a  Mixed  Number  to  within  a  certain  De- 
cimal  , 373 

Approximate  wth  Root  of  Numbers  entirely  decimal 374 

PERMUTATIONS  AND  COMBINATIONS 374 

Combinations 377 

Permutations  in  which  Letters  are  repeated 385 

Partial  Permutation 386 

General  Examples  in  Permutations  and  Combinations 387 

LOGARITHMS , 390 

First  Property.  The  Logarithm  of  the  Product  of  any  Number  of 
Factors  taken  in  the  same  System,  is  equal  t-o  the  Sum  of  the 
Logarithms  of  those  Factors 392 

Second  Property.  The  Logarithm  of  the  Quotient  arising  from  divi- 
ding one  Quantity  by  another,  is  equal  to  the  Logarithm  of  the 
Dividend  minus  the  Logarithm  of  the  Divisor 393 

Third  Property.  The  Logarithm  of  the  Power  of  a  Number  is  equal 
to  the  Exponent  of  the  Power  into  the  Logarithm  of  the  Number..  394 

Fourth  Property.  The  Logarithm  of  the  Root  of  a  Number  is  equal 
to  the  Logarithm  of  the  Number  divided  by  the  Index  of  the  Root.  395 

Fifth  Property.  The  Logarithm  of  the  Reciprocal  of  a  Number  is 
equal  to  the  Logarithm  of  the  Number  taken  negatively 396 


CONTENTS.  Xl 

PAGE 

LOGARITHMS  (continued). 

Sixth  Property.  The  Logarithm  of  any  Base,  taken  in  its  own  Sys- 
tem, is  Unity 397 

Seventh  Property.  If  we  have  a  Table  of  Logarithms  calculated  to  a 
Particular  Base,  the  Logarithms  of  the  Numbers  in  this  Table, 
divided  by  the  Logarithm  of  a  Second  Base,  taken  in  the  First 
System,  will  give,  as  Quotients,  the  Logarithms  of  the  same  Num- 
bers in  the  Second  System 397 

Differential  of  ax  398 

Logarithmic  Series 399 

Common  and  other  Logarithms  found  from  Napierian 401 

Measure  of  any  Modulus 401 

Table  of  Napierian  Logarithms 401 

Advantages  of  the  Common  System 404 

Application  of  Common  Logarithms 406 

To  find  the  Logarithm  of  a  given  Number  below  10000 407 

To  find  the  Logarithm  of  a  Number  above  10-000 408 

To 'find  a  Number  corresponding  to  a  given  Logarithm 409 

Solution  of  Exponential  Equations  by  means  of  Logarithms 411 

ARITHMETICAL  PROGRESSION 412 

Formula  for  the  nth  Term 413 

Formula  for  the  First  Term 414 

Formula  for  the  Number  of  Terms , 415 

Formula  for  the  Sum  of  the  Terms 41G 

To  find  the  Arithmetical  Mean 418 

Insertion  of  Means  between  the  Extremes  of  a  Proportion 420 

GEOMETRICAL  PROGRESSION 422 

Formula  for  the  nth  Term 423 

Formula  for  the  Ratio  of  the  Progression 424 

Formula  for  the  Sum  of  the  Series .' 425 

An  Infinite  decreasing  Progression 427 

INEQUALITIES 431 


Xll  CONTENTS. 

FAOE 

GENERAL  THEORY  OF  EQUATIONS 436 

Equal  Values 459 

Derived  Polynomials 462 

Equation  of  Differences 467 

Irrational  Values 470 

Newton's  Method  of  Approximation 474 

Lagrange's  Method  of  Approximation 475 

General  Solution  of  Numerical  Equations 477 

STURM'S  THEOREM 478 

DESCARTES'  RULE 491 

ELIMINATION  BETWEEN  TWO  EQUATIONS  OF  ANY  DEGREE 495 


ELEMENTS  OF  ALGEBRA 


Article  1. — A  mathematical  principle  is  a  truth  admitted  as  self- 
evident,  or  proved  by  a  course  of  reasoning  called  a  demonstration. 
Thus,  it  is  a  self-evident  principle,  that  if  equals  be  added  to  equals, 
the  results  will  be  equal.  But  it  requires  a  demonstration  to  show  that 
if  the  sura  of  two  quantities  be  added  to  the  difference  of  those  quanti- 
ties, the  result  will  be  equal  to  twice  the  greater  quantity. 

2.  Science  is  knowledge  gained  by  theory  or  experiment,  employed 
in  the  investigation  of  principles.  Art  is  the  application  of  acquired 
principles  to  the  practical  purposes  of  life. 

3.  Quantity  is  anything  that  can  be  measured  or  numbered.  A 
thing  is  said  to  be  measured  when  its  magnitude  is  expressed  in  terms 
of  the  unit  of  measure,  and  is  said  to  be  numbered  when  this  unit  is 
unknown  or  indefinite. 

Thus,  12  bushels  of  corn  is  a  definite  measure,  the  unit  of  measure 
being  one  bushel  of  corn.  But  a  quantity,  expressed  by  the  number 
12,  or  even  by  12  bushels,  would  only  be  numbered,  for  it  might  be  12 
bushels  of  gold,  or  12  of  flour,  or  12  of  anything  else. 

Lines  are  quantities,  becaiise  they  can  be  measured  in  terms  of 
yards,  feet,  inches,  &c.  Time  is  quantity,  being  measured  by  hours, 
minutes,  seconds,  &c.  The  unit  of  measure  for  time  is  generally  the 
second. 

The  operations  of  the  mind,  such  as  hope,  fear,  joy,  grief,  &c.,  are 
not  quantities.  For,  although  we  speak  of  a  great  hope  and  a  small 
hope,  there  is  no  definite  unit  of  comparison  by  which  to  measure  its 
magnitude. 

Numbers  are  not  quantities,  but  simply  the  agents  by  which  to 
express  the  abstract  relations  between  quantities  of  the  same  kind; 
that  is  to  say,  every  number  may  be  regarded  as  the  quotient  arising 
2  13 


14  ELEMENTS     OP    ALGEBRA. 

from  dividing  one  quantity  by  another  of  the  same  kind,  independently 
of  tlie  species  to  which  they  belong. 

Thus,  the  number  12  may  express  the  abstract  relation  between  12 
inches  and  1  inch,  or  between  12  months  and  1  month;  and  may,  there- 
fore, represent  the  quotient  arising  from  the  division  of  the  foot  by  the 
inch,  or  the  year  by  the  month.  And  so  it  may  express  the  quotient 
between  any  two  quantities  whatever,  provided  they  are  of  the  same 
kind. 

4.  Mathematics  is  the  science  of  quantity. 

Arithmetic  is  that  branch  of  mathematics  in  which  the  quantities 
considered  are  represented  by  numbers. 

5.  The  word  Algebra  is  of  Arabic  origin,  and  signifies  to  consolidate. 

Algebra  enables  us  to  investigate  arithmetical  principles  in  a  consoli- 
dated, and,  at  the  same  time,  general  manner,  and  may  be  regarded  as 
a  compendious,  and  also  universal  arithmetic. 

6.  The  quantities  considered  in  Algebra  are  represented  by  numbers 
and  letters,  and  the  operations  to  be  performed  are  indicated  by  signs. 
The  numbers,  letters,  and  signs  are  generally  called  symbols. 

The  numbers  and  letters  that  represent  quantities  are,  for  conve- 
nience, most  usually  called  quantities  themselves.  The  student,  how- 
ever, should  remember  that  they  are  only  the  representatives  of 
quantities. 

7.  The  sign  +  is  called  plus,  i.  e.,  more,  and  when  prefixed  to  a 
quantity,  signifies  that  it  is  to  be  added  to  some  other  quantity  expressed 
or  understood.  Thus,  a  +  6  is  read  a  plus  h,  and  signifies  that  h  is  to 
be  added  to  a.  The  expression  +  c  signifies  that  c  is  to  be  added  to 
some  quantity  not  expressed.  When  no  sign  is  written  before  a  quan- 
tity, the  sign  -f  is  always  understood.  In  the  expression  a  -f  &,  a  is 
understood  to  be  afiiected  with  the  plus  sign. 

8.  A  horizontal  line,  thus,  —  is  called  minus,  i.  e.  less,  and  when 
prefixed  to  a  quantity,  signifies  that  it  is  to  be  substracted  from  some 
other  quantity,  expressed  or  understood. 

Thus,  a  —  6  is  read  a  minus  h,  and  signifies  that  h  is  to  be  taken 
from  a.  The  expression  —  c  signifies  that  c  is  to  be  taken  from  some 
quantity  not  expressed. 

9.  A  Greek  cross  x  is  called  the  sign  of  multiplication,  and  when 
placed  between  quantities,  indicates  that  they  are  to  be  multiplied 
together.  Thus,  a  X  h  is  read  a  multiplied  by  h,  or  simply  a  into  h, 
the  sign  indicating  a  multiplication  to  be  performed. 


ELEMENTS     OE    ALaEBRA.  15 

10  The  multiplication  of  quantities  is  sometimes  indicated  by  a 
point.  Thus,  a  .  h  indicates  that  a  and  h  are  to  be  multiplied 
together. 

11.  The  multiplication  of  literal  factors  is  usually  indicated  by  wri- 
ting them  one  after  another,  thu^  a  i  c  is  the  same  as  a  X  6  X  c.  This 
notation  cannot  be  employed  when  numbers  are  used,  for  the  product 
thus  expressed  would  be  confounded  with  some  other  number.  The 
multiplication  of  2  by  4,  for  instance,  cannot  be  indicated  by  writing 
the  one  after  the  other,  because  the  product  would  be  mistaken  for  42 
or  24. 

12.  When  several  terms  connected  by  the  sign  +,  or  — ,  arc  to  be 
multiplied  by  a  single  term,  the  multiplication  is  indicated  by  means 
of  parentheses.  Thus,  (a  -\-  h  -{■  c)  m,  signifies  that  the  sum  of  a,  b, 
and  c  is  to  be  multiplied  by  m.  "When  the  multiplier  itself  is  composed 
of  more  than  one  term,  it  is  also  enclosed  in  parenthesis:  Thus,  (a  +  b) 
(m  —  c)  indicates  that  the  sum  of  a  and  b  is  to  be  multiplied  by  the 
diflFerence  of  m  and  c.  A  horizontal  or  vertical  line  is  also  used  to  col- 
lect terms  for  multiplication.  Thus,  a  x  to  +  «  -f  c,  indicates  that  the 
sum  of  m,  n,  and  c  is  to  be  multiplied  by  a.  The  same  thing  may  be 
indicated  by  a  vertical  bar,  thus 

4-TO  I 

+  n 
+  c 

13.  The  coefficient  of  a  quantity  is  a  number  or  letter  prefixed  to  a 
quantity,  showing  how  often  it  has  been  added  to  itself.  Thus,  instead 
of  writing  a  +  a  -r  a,  which  denotes  the  addition  of  a  three  times,  we 
abridge  the  notation  by  writing  3a.  So  also,  10.ry,  signifies  the 
addition  of  xj/  ten  times.  In  like  manner,  mx  signifies  the  addition 
of  X,  m  times.  The  coefficient  serves  as  a  brief  mode  of  indicating  the 
addition  of  a  quantity  to  itself.  When  no  coefficient  is  written,  1  is 
always  to  be  understood. 

14.  The  exponent  is  a  small  number  or  letter  written  a  little  above 
and  to  the  right  of  a  number  or  letter,  and  indicates  the  number  of 
times  it  enters  into  itself  as  a  factor.     Thus,  we  write 

a^  instead  of  aa,    and  call  the  result  a  square. 

a^      "       "  aaa,    "      "     "       "      a  cube. 

a*      "       "  aaaa, "      "     "       "      a  to  the  fourth  power. 

The  exponent  enables  us  to  abbreviate  the  manner  of  indicating  the 


16  ELEMENTS     OF    ALGEBRA. 

multiplication  of  a  quantity  by  itself.  When  no  exponent  is  wiitten,  1 
is  always  to  be  understood. 

Division  is  denoted  by  three  signs.     The  division  of  a  by  &  may  be 

indicated  hj  a-i-h,  or  — ,  or  a\b. 

15.  The  sign  =  is  called  the  sign  of  equality,  and  is  read  "  equal 
to."  When  placed  between  two  quantities,  it  indicates  that  they  are 
equal  to  each  other.  Thus,  a  =  b,  is  read  a  equal  to  b.  In  like  man- 
ner, 2  +  4  =  6  is  read  2  plus  4  equal  to  6. 

16.  The  sign  ^  is  called  the  sign  of  inequality,  and  is  read  "greater 
than,"  when  the  opening  is  toward  the  left;  and  "less  than,"  when 
opening  is  toward  the  right.  Thus,  a  ^  Z>  is  read,  a  greater  than  b ; 
and  c  <^  m  is  read,  c  less  than  m. 

17.  A  root  of  a  quantity  is  a  quantity  which,  multiplied  by  itself  a 
certain  number  of  times,  will  produce  the  given  quantity.  To  indicate 
the  extraction  of  a  root,  we  use  the  sign  V ,  called  the  radical  sign, 
placing  a  number  or  letter  to  the  left  and  over  the  sign  to  indicate  what 
root  is  to  be  extracted.  Thus,  ^'a  denotes  the  square  root  of  a; 
%/~a  the  cube  root  of  a ;   V~a  the  i\>'^  root  of  a. 

18.  The  number  or  letter  placed  over  the  radical  sign  is  called  the 
index  of  the  radical.  When  no  index  is  written,  we  always  understand 
that  the  square  root  is  to  be  extracted.  Thus,  s/  a  means  that  the 
square  root  of  a  is  to  be  taken. 

19.  A  simple  quantity  is  one  in  which  the  letters  and  numbers  of 
which  it  is  composed  are  not  connected  by  the  sign  plus  or  minus. 

Thus,  a,  aZ>,  —  and  c  are  simple  quantities  or  monomials.  All  quan- 
tities not  simple  are  compound,  and  called  binomials  when  composed 
of  two  terms;  trinomials  when  composed  of  three;  and  polynomials 
when  composed  of  more  than  three.  Each  of  the  literal  factors  which 
enter  into  a  term  is  called  a  dimension  of  the  term.  The  degree  of  a 
term  is  the  number  of  its  dimensions.  When  a  factor  enters  more  than 
once,  its  dimension  is  denoted  by  the  exponent.  Thus,  a^  is  of  the 
second  dimension. 

20.  When  several  factors  are  multiplied  together,  the  sum  of  the 
exponents  denotes  the  dimension  of  the  term.  Thus,  ab  is  of  the 
second  dimension,  af-bc,  of  the  fourth,  aWcd,  of  the  sixth,  &c.  If 
the  term  is  a  fraction,  its  degree  is  denoted  by  the  difference  between 


ELEMENTS   OF   ALGEBRA.  17 

the  sums  of  the  exponents  of  the  numerator  and  denominator.     Thus 
is  of  the  second  degree,  — ~  of  the  first  degree,  &c. 

21.  A  monomial  is  said  to  be  liomogcneoxis,  ^hen  all  of  its  literal 
factors  are  of  the  same  dimension.  A  polynomial  is  said  to  be  homo- 
geneous, when  all  its  terms  arc  of  the  same  degree.  Thus  ah,  a?h^, 
a!^L^,  are  each  separately  homogeneous  monomials  ;  4a*Z/  -f  2Z/V  —  m^,  is 
a  homogeneous  polynomial. 

22.  Like  quantities  are  those  which  are  composed  of  the  same  letters, 
and  which  have  their  corresponding  letters  affected  with  the  same  ex- 
ponents ;  SaP  +  lOah^  —  3a6^  are  like  quantities.  But  Sai-  -f  lOah 
—  Sa^b'^  are  unlike ;  for,  though  the  letters  are  like,  the  exponents  of 
these  letters  are  different.  p.  17. 

23.  The  reciprocal  of  a  quantity  is  1  divided  by  that  quantity.     The 

.1  1  2    13 

reciprocal  of  2  is  -- ;  of  a,  - ;  of  r  ■  o  o^  o>  •-'*^'^- 
2  a  3    2       2 

3 

24.  Quantities,  affected  with  the  plus  i?ign,  are  called  iwsitive  quanti- 
ties, and  those  affected  with  the  minus  sign,  are  called  ncfjative  quan- 
tities. The  student,  however,  should  bear  in  mind  that  quantities  can- 
not be  positive  or  negative  in  themselves,  and  that  by  these  phrases,  we 
wish  merely  to  signify  additive  and  subtractivc  quantities. 

25.  To  familiarize  the  student  with  the  foregoing  symbols,  we  sub- 
join a  few  examples  fur  practice  :  — 

1.  Express  in  algebraic  language  that  twice  the  product  of  x  into  y, 
divided  by  t^iree  times  the  product  of  z  into  to,  shall  be  equal  to  G. 

2.  Express  that  the  sum  of  a  and  b,  when  added  to  their  difference 
is  equal  to  twice  the  greater  quantity  a. 

3.  That  one-third  of  a  quantity  ??i,  multiplied  by  ith  of  a  second 
quantity  ?),  and  that  product  increased  by  100,  the  result  will  be  a  hun- 
dred thousand. 

4.  Find  the  value  of  this  expression, ,  when  b  =  G-i:,  f  =  144, 

m.n 

m=10,  n=l. 

Find  the  value  of  the  expression (x  a),  when  m  =  4,  ?i  =  3, 

x  =  2,  a  =  l. 

Find  the  value  of  the  expression  —,-,  when  7?i  =  4,  n  =  S,  c:^=6, 

nrii 
h  =  10,  1  =  100. 

2*  B 


18  ADDITION. 

ADDITION. 

26.  Addition  is  the  connecting  of  several  terms  together  by  the 
sign  plus  or  minus,  so  that  they  nuiy  be  reduced  into  a  single  expres- 
sion. 

There  are  three  distinct  cases  :  — 

CASE  I. 

27.  When  the  terms  are  like  and  have  like  signs. 


Add  the  coefficients  of  the  several  terms  together,  prefix  the  common 
sign  to  this  sum,  and  write  after  it  the  common  letter  or  letters,  with 
their  primitive  exponents. 

Thus,  +  2a  +  3a  are  like  quantities  referred  to  the  same  unit,  and 
may  therefore  be  added,  just  as  two  whole  numbers  are  added  in  Arith- 
metic. The  -1-  2a  indicates  that  a  is  to  be  added  twice  to  some  quan- 
tity not  expressed,  and  +  3a  that  a  is  again  to  be  added  three  times  to 
the  same  quantity.  This  addition  can  obviously  be  indicated  at  once 
by  writing  +  5a  instead  of  +  2a  4-  3a.  Suppose,  for  example,  that  a 
represented  one  dollar,  then  2a  would  represent  two  dollars,  and  3a 
three  dollars,  and  the  sum,  five  dollars,  would,  of  course,  be  represented 
by  5a.  In  like  manner,  — 36  —  46  is  equal  to  —  76,  because  the  minus 
signs  indicate  that  the  quantities  represented  by  36  and  46  are  to  be 
taken  from  some  quantity  not  expressed,  and  subtracting  76  at  once  is 
obviously  the  same  as  first  subtracting  36  and  then  subtracting  46.  We 
may  then  write  — 36  —  46  =  —  76,  the  minus  sign  before  the  76  indi- 
cating that  it  is  to  be  taken  from  some  quantity  not  expressed,  and  the 
whole  expression  denoting  that  the  subtraction  of  76  is  the  same  as  the 
successive  subtraction  of  36  and  46.  This  will  be  made  plainer  to  the 
beginner  by  attributing  to  6  some  known  value,  as  a  pound  or  an  ounce. 


EXAMPLES. 

-f      a 

—     a 

+     &—    y 

+      m —     z 

—  4,-2 

+  2f 

+    2a 

—   2a 

+    26-  2^ 

+   5„i_   2s 

—  5z' 

+  5/ 

-f-    3a 

—   3a 

-f    56 —  4y 

+    8??!,—   3z 

—  9,r 

+   7f 

-1-    4a 

—  4a 

+    96—  8y 

+  10m—    7z 

—  12,-2 

+  sy 

+"io^" 

~10^' 

+  176—15^ 

+  2im  —  ldz 

—  W^ 

+  22/ 

ADDITION.  19 

CASE  II. 

28,  When  the  quantities  are  like,  but  affected  with  unlike  signs. 

RULE. 

Add  the  quantities  affected  tcith  the  positive  siyn  hy  the  last  ride,  then 
add  those  affected  with  the  negative  sign  in  like  manner.  Subtract  the 
smaller  of  the  coefficients  of  those  sums  from  the  greater,  annex  the 
common  letter  or  letters  to  the  difference,  and  prefix  the  sign  of  the 
greater  sum. 

If  we  were  required  to  add  -{-  ba  —  60  +  4a  —  a  together,  we  could, 
for  the  reasons  already  given,  write  +  9a  for  the  positive  terms,  and  7a 
for  the  negative  terms.  By  this,  we  must  understand  that  9a  has  to  be 
added  to  some  quantity  not  expressed,  and  that  7a  has  to  be  subtracted 
from  the  sum  of  9a,  and  this  unexpressed  quantity.  Now  to  take  7a 
from  9a,  and  add  the  remainder  to  the  unexpressed  quantity  is  plainly 
the  same  as  taking  7a  from  the  sum  of  9a,  and  the  unexpressed  quan- 
tity. But  the  difference  between  9a  and  7a  is  2a,  hence  the  sum  of 
-f  5a  —  6a  +  4a  —  a  is  +  2a.  The  plus  2a  denotes  that  2a  has  to 
be  added  to  an  unexpressed  quantity. 

If  we  were  required  to  add  —  5a  +  Ga  —  4a  -f-  a  together,  we  could, 
as  has  been  shown,  collect  the  negative  terms  into  a  single  term,  —  9o, 
and  the  positive  terms  into  a  single  term  -|-7a.  Wc  would  then  be 
required  to  take  9a  from  the  sum  of  7a  and  some  unexpressed  quantity, 
or,  which  would  be  the  same  thing,  to  take  9a  from  7a,  and  add  the 
remainder  to  the  unexpressed  quantity.  But  9a  is  made  up  of  7a 
added  to  2a,  and  when,  therefore,  we  subtract  9a  from  7a,  the  7a's 
will  strike  each  other  out,  and  there  will  still  remain  2a  to  be  sub- 
tracted. The  required  and  unexecuted  subtraction  is  indicated  by  the 
minus  sign  before  the  2a,  and  the  true  sum  of  —  5a  -f  6a  —  4a  -f  a 
is  therefore  —  2a. 

EXAMPLES. 

1.  Add  2  1  — Gh  +  ah  — 27j  together. 

Ans.  +ah  —  6i  or  (+  a  —  6)  J. 

2.  Add  4c  -{-  3c  —  mc  -{-  no  together. 

Ans.  (7  -\-  71  —  w)  c. 

3.  Add  xi/  -{-  2xy  +  hxy  — pxy  together. 

Ans.  (3  -f  &  — p)  xy. 


20  ADDITION. 

4.  Add  4a  —  7a  —  3a  —  10a  —  12a  together. 

Ans.  —  28a. 

5.  Add  4ax  —  7ax  —  dax  —  Wax  —  12acc  together. 

A71S.  —  2Sax. 

CASE  III. 

29.  When  the  quantities  arc  similar  and  dissimilar,  and  have  like 
and  unlike  signs. 

RULE. 

Add  all  (he  sets  of  similar  terms  hy  the.  last  two  rules,  and  write 
their  sums  one  after  another,  and  connect  with  them  all  the  single  terms 
with  their  oion  signs. 

Since  the  letters  are  always  the  representatives  of  quantities,  it  is 
obvious  that  the  quantities  represented  by  dissimilar  terms  cannot  be 
added  into  one  sum.  Thus,  4a  +  46  neither  make  8a  nor  86.  The 
quantity  a  might  represent  a  year,  and  the  quantity  h  a  pound ;  then 
4a  would  represent  4  years,  and  Ah  would  represent  4  pounds,  and  the 
addition  ought  neither  to  give  8  years  nor  8  pounds.  Hence,  we  can 
only  reduce  the  similar  terms  in  sets,  and  connect  the  results  with  their 
appropriate  signs. 


3jn^  —  n^    xy  +  Ix        bax  +  y        2nx+  w 

3x^  +  ni'  z  -\-y  x^  -\-   Ay      2w  -\-  bnx 

2m^  -\- x^    xy  +ox        Ax}  +    %       6».x4-s 


Qm^  +  Ax^  —  n""    2xy  +  Wx+z^y    bx^-\-bax^lAy    V6nx  +  ^io-\-s 

30.  It  often  facilitates  an  algebraic  operation  to  arrange  the  terms  in 
a  certain  order.  In  addition,  this  arrangement  is  effected  by  placing 
all  the  like  terms  beneath  one  another. 

Ta^  4  Qxy  +  5^    +      Aw  la"  +    Aw  +    Qxy  +    bz 

w  -\-Az    +  8a-y+     12a^  may  be  written    12a- +      ?r  +     'ixy  +    Az 

Sz   +  9xy+  bw  +  100a-  100a"  +    bw  +    9xy  +     Sz 

llda'+10w  +  2nxy  +  Viz 

31.  Sometimes  the  addition  of  a  compound  quantity  to  a  single  quan- 
tity, or  to  another  compound  quantity,  is  not  actually  performed,  but 
indicated  by  the  parenthesis  (),  or  vinculum.  Thus,  46 +  (6 — c),  or 
4t  ^  Ij  —  c,  indicate  that  h  —  c  is  to  be  added  to  Ah. 


SUBTRACTION.  21 


Remarks. 

32.  It  will  be  noticed  tliat  the  term  addition  in  Algebra,  is  used  in 
an  extended  sense,  the  operation  being  often  arithmetical  substraction. 
The  addition  of  a  negative  quantity  is,  in  fact,  the  same  as  the  subtrac- 
tion of  the  same  quantity  regarded  as  positive.  The  use  of  negative 
quantities  in  Algebra  constitutes  one  of  its  marked  drfiferences  from 
Arithmetic,  in  which  the  numbers  employed  are  always  supposed  to  be 
positive.  By  reference  to  the  examples,  another  remarkable  distinction 
between  Algebra  and  Arithmetic  will  be  observed ;  each  sum  indicates 
Avhat  (|uantities  were  added  together,  whilst  an  arithmetical  sura  con- 
tains no  trace  of  the  numbers  employed.  Thus,  the  sum  of  10  and  5 
is  15,  but  the  result  does  not  point  out  the  numbers  that  were  added, 
for  15  might  proceed  from  the  addition  of  14  and  1,  12  and  o,  13 
and  2,  &c. 

It  will  be  seen  hereafter,  that  in  all  algebraical  operations,  the 
result  contains  some,  if  not  all,  the  quantities  employed  in  the  inves- 
tij^atious. 


SUBTRACTION. 

33.  Subtraction  is  taking  the  dlfR'rence  between  two  or  more 
quantities,  and  may  be  regarded  as  the  undoing  of  a  previous  addition. 

If  we  were  required  to  subtract  +  G  —  4,  or  2  from  12,  the  result 
plainly  ought  to  be  10.  But  if  we  write  G  —  4  beneath  the  12,  and 
perform  the  subtraction  of  each  term  separately,  the  subtraction  of  6 
from  12  will  give  a  remainder  G,  which  is  too  small  by  4,  because  we 
were  not  re^juired  to  take  6,  but  2  from  12.  We  can  only  correct  the 
error  by  adding  +  4,  and,  hence,  we  have  12  —  (  +  G  —  4)  =  12  — 
G  4-  4  =  10,  as  before.  We  observe,  in  the  last  result,  that  the  signs 
of  G  and  4  have  both  been  changed. 

Again,  let  it  be  required  to  subtract  +  i  —  c  from  a.  The  sub- 
traction of  h  from  a  will  be  indicated  by  writing  a  —  h,  but  tliis  re- 
mainder is  too  small  by  c,  because  we  were  not  required  to  subtract  h 
itself  from  a,  but  what  remained  of  h  after  it  was  reduced  by  c.  The 
error  can  only  be  corrected  by  adding  plus  -:  to  the  result.  Hence, 
a  —  (+Z>  —  c)=  a  —  6  +  c. 


SUBTRACTION, 


In  this  general  example,  we  see  that  all  the  signs  of  the  subtrahend 
have  been  changed,  and  then,  it  has  been  added,  as  in  addition. 
Hence,  we  have  the  following 


RULE. 


Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to  he  changed, 
and  then  add  up  these  terms  as  in  addition. 


EXAMPLES. 

—   6a  +  46 

Sy/~X   +    5a2 

7x^  -r    m 

2  V  a  +     ax 

+    4a—    i 

Q^'x—  Sa^ 
—  oVx  +  loa- 

7x'  +  2m 
0  —     m 

4s/of  +  Qax 

—  10a  +  5b 

—  2Va  —  bax 

From  y/x  +    b  —  10   +Sa^+7Qi/   +18a+ 19x  +  lira 
Take       d —20  +    dx  +  bn—2^x+    36+    4aH.:;  +  2s 

3v/x-  — 26  +  10*  +4a^  +  lSa+Qn—d—z—2s  +7Qi/  +  10 

In  the  third  example  we  have  written  zei'o  as  the  difference  between 
7x^  and  7x^,  but  it  is  more  usual  to  denote  zero  by  a  blank. 

From    iixt/  +  14  v/y  +    s         +  2n  —    w 
Take    Sxi/  +  2  —  11  +  11  v/y—  2n  +     w 

11         +    SV7+  4n  — 2m; 

From  12a^  +  Ux  —  cx^  +  m^  +  46-  —  c^x^  +  71s 
Take     46^  —  cV  —  Qbx  +  ex''  —  m!"  —  3«s  +  106^  —  io>/  +  12a- 
106x  — 2cx^  +  2m^  +  4ns  — 10b-  +  wi/  " 

34.  The  remainder  added  to  the  subtrahend  ought  to  be  equal  to  the 
minuend,  and  the  result,  therefore,  can  be  verified,  as  in  Arithmetic. 

It  will  be  seen  that  an  Algebraic  subtraction  does  not  necessarily  im- 
ply diminution,  and  that  the  remainder  may  not  only  exceed  the  sub- 
trahend and  minuend  separately,  but  may  be  equal  to  their  arithmetical 
sum.  In  general,  the  subtraction  of  a  negative  quantity  is  equivalent 
to  the  addition  of  the  same  quantity  taken  positively.  Thus,  —  6  taken 
from  a  gives  a  +  6  for  a  remainder.  This  subject  may  be  illustrated 
by  a  simple  example :  Suppose  a  to  represent  the  value  of  an  estate 
exclusive  of  its  liabilities,  and  6  a  mortgage  upon  that  estate,  then  a  —  b 


SUBTRACTION.  23 

will  represent  the  actual  value  of  the  estate.  Now,  suppose  some  friend 
of  the  owner  should  determine  to  cancel  his  debt ;  he  could  do  this, 
either  by  removing  the  mortgage  (in  algebraical  language,  taking  away 
—  h),  or  by  giving  him  a  sum  of  money  equal  to  the  debt,  that  is, 
a  sum  represented  by  +  h.  So  we  see  that,  taking  a-^vay  —  6  is  equi- 
valent to  adding  -f-  h. 

35.  Quantities  arc  considered  negative,  when  opposed  in  character  or 
direction  to  other  quantities  of  the  same  kind,  that  are  assumed  to  be 
positive. 

Thus,  if  a  ship  leave  port  with  the  intention  of  sailing  due  north,  but 
encounters  adverse  winds,  and  is  driven  south  part  of  the  time;  to  get 
the  distance  passed  over  .north,  we  must  obviously  subtract  the  distance 
sailed  over  south.  If  then  the  direction  north  be  considered  positive, 
the  direction  south  must  be  considered  negative.  Suppose  the  vessel 
sails  first  day  100  miles  north  to  10  south,  second  day  80  miles  north 
and  30  south,  the  entire  distance  passed  over  in  a  northern  direction 
will  be  expressed  by  100  +  80  — 10  — 30  =  140. 

If  a  man  agree  to  labor  for  a  dollar  per  day,  and  to  forfeit  half  a  dollar 
for  every  idle  day,  and  he  labor  4  days  and  is  idle  2,  his  wages  will  be 
4x  1  —  2x  J,  or4x  l  +  2( —  *)  =  3.  We  see  that  we  have  re- 
garded as  negative,  cither  the  forfeiture  as  opposed  to  gain,  or  the  idle 
days  as  opposed  to  the  working  days. 

If  a  man's  age  be  now  thirty  j-ears,  his  age  four  years  hence  will  be 
expressed  by  30  +  4 ;  his  age  four  years  ago  by  30  —  4.  Here  future 
time  being  positive,  past  time  is  negative.  If  a  steamboat  sail  with  a 
velocity  of  10  miles  per  hour,  and  encounter  a  head  wind  that  would 
carry  it  back  at  the  rate  of  8  miles  per  hour,  then  its  rate  of  advance 
will  be  expressed  by  10  —  8,  or  2  miles  per  hour.  But  if  it  be  carried 
back  at  the  rate  of  12  miles  per  hour,  then  its  rate  of  advance  will  be 
expressed  by  10  — 12,  or  —  2  miles  per  hour.  It  will  then  plainly  be 
carried  back,  and  the  minus  indicates  a  change  of  direction. 

Distance,  when  estimated  as  positive  in  one  direction,  is  considered  nega- 
tive in  the  opposite  direction.  Thus,  let  the  distance  AB  =  n  and  BC  =  w, 

C 

'- i ,  the  distance  BC  being  considered  positive  on  the 

A         B  '      '  °  ^ 

right  of  B.  Then  AC  ^  wi  +  n.  Now  suppose  the  distance  BC  be  esti- 
mated on  the  left  of  B,  the  point  C  falling  between  B  and  A,  then 
AC  =  m  —  n.  We  see  that  the  distance  BC,  which  was  regarded  as 
positive  on  the  right  of  B,  became  negative  when  estimated  on  the  left. 


24  MULTIPLICATION. 

In  general,  the  minus  sign  may  be  regarded  as  always  indicating 
either  that  a  quantity  has  changed  its  character  or  direction,  or  that  it 
is  just  the  opposite  of  something  else  of  the  same  kind  that  is  con- 
sidered positive. 


MULTIPLICATION. 

36.  Multiplication  is  a  short  method  of  adding  the  multiplicand 
to  itself  as  many  times  as  there  are  units  in  the  multiplier.  Thus,  to 
multiply  a  by  Z)  is  to  add  a  to  itself  h  times,  and  since  the  addition  of 
a  to  itself  twice  is  2a,  three  times  3a,  &c.,  the  addition  of  a  to  itself  h 
times  will  be  ha.  Hence  the  product  of  a  by  &  will  be  la  or  ah,  for  it 
plainly  matters  not  in  what  order  we  write  the  factors. 

There  are  three  cases  :  — 


CASE  I. 

37.  When  both  multiplicand  and  multiplier  are  simple  quantities. 

Before  giving  a  rule  for  the  multiplication,  it  will  be  necessary  to 
show  that  the  product  of  simple  quantities  having  like  signs,  both  plus 
or  both  minus,  is  always  positive,  and  that  the  product  is  always  nega- 
tive when  they  have  unlike  signs.  The  product  of  +  a  by  -{-h  is 
plainly  +  ah,  because  plus  a  added  to  itself  h  times  m.ust,  of  course,  be 
positive. 

And  —  a  by  -\-  h  must  be  negative,  because  —  a  added  to  itself  any 
number  of  times  must,  of  course,  retain  its  sign.  But  — a  by — h 
will  give  +  ah,  for  the  h  having  a  minus  sign  before  it,  indicates  the 
reverse  of  what  it  did  before,  and  therefore  denotes  that  —  a  is  to  be 
subtracted  from  itself  h  times.  But  we  have  seen  that  the  subtraction 
of  —  a  is  the  same  as  the  addition  of  H-  a  ;  in  like  manner,  the  subtrac- 
tion of  —  a  twice  is  the  same  as  the  addition  of  +  2o  ;  the  subtraction 
of  —  a  tliroe  times  the  same  as  the  addition  of  +  3a  ;  and  the  subtraction 
of  —  a,  h  times,  the  same  as  the  addition  of  +  a,  +  h  times,  or  +  ah. 
Hence,  the  product  —  o  by  —  &  is  +  ah,  and  we  see  that  like  signs 
produce  plus,  and  unlike  signs,  minus. 

Let  it  now  be  required  to  multiply  a^  by  a".  The  exponents  indicate 
that  a  enters  twice  as  a  factor  in  a^,  and  four  times  as  a  factor  in  a, 
hence,  it  will  enter  6  times  as  a  factor  in  the  result. 


MULTIPLICATION.  25 

Hence,  a^  x  a''  :=  aaaaaa  =  a^,  and  we  see  that  the  multiplication  is 
effected  by  adding  the  exponents  of  the  same  letter. 

To  multiply  a^  by  ab  is  the  same  as  multiplying  a^  by  a,  and  then 
multiplying  the  result  by  h.  But  to  multiply  a^  by  a,  we  have  only  to 
add  the  exponents  of  a^  and  a,  hence  the  result  is  a'.  Now  multiply 
by  b,  and  we  have  a'^b. 

The  multiplication  is  effected  by  adding  the  exponents  of  like  letters, 
and  writing  after  the  result  the  letter  which  is  not  common  to  the  mul- 
tiplicand and  multiplier,  with  its  primitive  exponent.  In  like  manner 
a^  X  ah^  will  be  a^b^.  To  multiply  a\-  by  ab,  is  the  same  as  multiplying 
a^c  by  a,  and  then  the  result  by  b.  But  a'^c  by  a,  as  we  have  seen,  gives 
a^c,  and  that  product  by  b,  will  plainly  be  a\b.  And  the  multiplica- 
tion is  again  effected  by  adding  the  exponents  of  like  letters,  and  writing 
after  the  result  the  letters,  which  are  not  common,  affected  with  their 
primitive  exponents.  The  same  reasoning  can  be  extended  to  any  num- 
ber of  factors,  the  multiplication  being  effected  in  every  instance,  by 
adding  the  exponents  of  like  letters,  and  writing  after  the  result,  the 
letters  not  common,  with  their  primitive  exponents.  Thus,  a^cd  by  ab 
gives  aW6,  and  ah-d  by  abm  gives  a\dbni.  We  have  taken  mono- 
mials whose  coefficients  were  unity.  If  we  were  required  to  multiply 
a  by  2b,  the  product  of  a  by  6  has  to  be  nuiltiplied  by  2,  but  the  pro- 
duct a  by  6  is  ab,  hence  the  result  is  2ab.  If  we  were  required  to 
multiply  2a  by  b,  then  2a  has  to  be  added  to  itself  as  many  times  as 
there  are  units  in  b,  but  to  add  2a  to  itself  twice  gives  2a  x  2  or  4a, 
to  add  it  three  times  gives  2a  X  3,  or  Ga,  and  to  add  it  b  times  gives 
2a  X  b,  or  2ab.  In  like  manner,  if  we  were  required  to  multiply  2rt 
by  4i,  then  2a  has  to  be  added  to  itself  -ib  times,  and  the  result  will 
plainly  be  2a  x  46,  or  Sab :  the  same  result  that  we  would  get  by 
multiplying  the  literal  factors  together,  and  prefixing  the  product  of 
the  coefficient  for  a  new  coefficient.  Hence  we  have  for  the  multipli- 
cation of  monomials,  the  following 

RULE. 

38.  Multqjlr/  tJie  coefficients  to<jctlicr  for  a  neio  coefficient,  and  write 
after  it,  all  the  literal  factors,  common  to  the  tico  monomials,  affected 
with  exponents  equal  to  the  sum  of  the  exponents  in  the  multiplicand 
and  midtipUer,  and  those,  ichich  are  not  common,  with  their  primitive 
exponents.  If  the  monomials  have  like  signs,  give  the  phis  sign  to  the 
product ;  if  they  have  xmlike  signs,  give  the  minus  sign  to  the  j^roduct. 
3 


MULTIPLICATION, 


EXAMPLES. 

1.  a'hc 

by  ahc. 

^ns.  a»iV. 

2.  —a%c 

by  a&c. 

^7is.  —  a^iV 

3.  a'hc 

by  — ahc. 

^Tis.  —  a'Z.^cf 

4.  2a%c 

by  aic. 

.4?;s.  2a»6V. 

5.  2a%c 

by  aic. 

Jns.  —  2a'^^c^ 

6.  2a%c 

by  — ahc. 

^«s.  —  2a='iV. 

7.  a^Sc 

by  2a  &c. 

^«s.  2a='iV. 

8.  d'hc 

by  —  2a5c. 

^jis.  —  2a%^ 

9.  a-'Z." 

bya"'+'i"  +  '. 

^72s.  a'^-^'i^"". 

10.  «  +  -&» 

by  a" -'?<"-' 

^«s.  a^^-'i^-'. 

11.  a-n'i-»c-» 

by  a^h-'c\ 

^jis.  a''Z.-"-2c-"   =>. 

12.  4«-'"i"c 

by  —  Sa^i-'c-^ 

^72s.  —  32a''^"-='c-3. 

13.  4a^6V 

by  —  Sa^i^C 

Jws.—  32a^Z;V. 

14.  6a-='i-''c--3 

by  7a-^i-V-''. 

Ans.  42a-^i-^c-l 

15.  3a«Z;V 

by  9a-='i-'^c-^ 

^rjs.  27a^Z-V. 

16.  a^^V 

by  Axyzh. 

Ans.  Ax^    Y+'zf+'h. 

17.  a^^^-rc 

by  ac. 

A71S.  +  x^yzac^. 

18.  x^fc 

by  xy. 

Ans.  x^    yc. 

19.  rr^/c 

by  a-7/^ 

Ans.  x^'^\ifc^ 

20.  x^-\jc 

by:.y. 

Ans.  x^^Yc 

21.  rr-'-yc-^ 

by  cV. 

Ans.  x^^yc. 

Remnrlcs. 

39.  We  see,  from  the  last  four  examples,  that  we  can  make  any 
change  in  the  position  of  the  exponents,  provided  we  retain  the  factors, 
and  get  the  same  result. 

We  see,  from  examples  13,  14,  and  15,  that  when  the  factors  of  the 
multiplicand  and  multiplier  are  homogeneous,  the  factors  of  the  result 
will  also  be  homogeneous. 

Example  11  shows  that  this  will  not  be  true  when  the  multiplicand  is 
alone  made  up  of  homogeneous  factors ;  and  Example  1  shows  that  it 
will  not  be  true  when  the  multiplier  alone  is  homogeneous. 

40.  Any  number  of  monomials  may  be  multiplied  together,  in  accord- 
ance with  the  preceding  principles. 


MULTIPLICATIOy  27 


EXAMPLES. 


1.  o\"  X       abX       hdXmn.  Ans.       a?l?dcmn. 

2.  — ah  X — ah  X       Id  X  nui.  Ans.        a^U^dcmn. 

3.  — a'c  X — ah  X  —  hd  X  mn.  Ans.  — a%\Jcmn. 

4.  — a-c  X — ah  X       hd  X  —  mn.  Ans.  — a^h^dcmn. 

We  see,  that  when  the  monomials  are  all  positive,  the  result  will  be 
positive ;  and  when  the  negative  monomials  are  even  in  number,  the 
result  will  also  be  positive.  But  when  an  odd  number  of  negative  mo- 
nomials are  multiplied  together,  as  in  the  fourth  Example,  or  when  an 
odd  number  of  negative  monomials  are  multiplied  by  a  positive  mono- 
mial, as  in  Example  3,  the  result  will  be  negative.  The  result  would 
also  be  negative,  if  an  odd  number  of  negative  monomial  factors  was 
multiplied  into  any  number  of  positive  monomial  factors. 


CASE   II. 

41.  When  tlae  multiplicand  is  a  compound  quantity  and  the  multi- 
plier a  monomial. 

The  multiplicand  is  to  be  repeated  as  many  times  as  there  are 
units  in  the  multiplier;  each  term  of  the  multiplicand  is  then  to  be 
multiplied  by  the  multiplier,  and  the  partial  products  to  be  connected 
with  their  appropriate  signs.  We  might  a.ssumc,  what  has  already 
been  proved  for  monomials,  that  the  product  of  a  positive  quantity  by  a 
positive,  or  of  a  negative  quantity  by  a  negative,  gives  a  positive  result; 
and  that  the  product  of  a  positive  quantity  by  a  negative  gives  a  nega- 
tive i-esult.  But  we  can  demonstrate  this  rule  more  rigorously  :  Let  it 
be  required  to  multiply  a  —  a  hy  +  c.  We  know  that  a  —  a  is  zero, 
and  that  zero  repeated  c  times  must  still  be  zero.  Hence,  the  product 
of  a  —  a  by  +  c  must  be  zero.  But  +  ahy  +  c  gives  -|-  ac  for  a 
product,  and,  therefore,  the  product  of  —  a  by  -|-  c  must  be  —  ac,  in 
order  to  cancel  the  first  product.  Hence,  the  product  of  a  negative 
quantity  by  a  positive  gives  a  negative  result.  Or,  to  express  the 
whole  algebraically, 

a   —  a    =  0 

+  c 

■j-ac  —  ac  =  o 


28  M  U  Ti  T  I  P  L  I  C  A  T I  0  N  . 

Let  it  now  be  re((uired  to  multiply  a  —  a  by  —  c.  The  multipli- 
cand being  zero,  tlie  product  must  be  zero.  But,  from  wliat  has  just 
been  shown,  the  product  of  +  a  and  —  c  is  —  ac,  hence,  the  product 
of  —  a  by  —  c  must  be  +  ac,  to  destroy  the  first  product.  Or,  in 
other  words,  the  product  of  a  negative  quantity  by  a  negative  must  be 
effected  with  the  positive  sign. 

Algebraically,         a   —  a  =  o 
—  c 


We  conclude  that  the  product  of  a  positive  quantity  by  a  positive,  and 
of  a  negative  quantity  by  a  negative,  is  positive ;  and  that  the  product 
of  a  positive  quantity  by  a  negative  is  negative.  Briefly,  we  say  like 
signs  give  a  positive  result,  and  unlike,  a  negative  result. 

RULE. 

Multiply  each  term  of  the  vudtiplicaud  hy  tlie  midliplicr,  and  con- 
nect the  res^dts  ivith  their  appropriate  signs. 


EXAMPLES. 

a^  -{-h                 by  a. 

Ans. 

a""  -f  ah. 

a^ — h                 by  a. 

A71S. 

a^~ah. 

—  a^  -\-h            by  a. 

Ans. 

—  a^  +  ah. 

—  d'  —  h            by  — 

a. 

Ans. 

«'  -f  o&. 

—  fr  -f  Z»            by  — 

a. 

Ans. 

a^'  —  ah. 

a^t  +  c  -f  d        by  a. 

Ans. 

a^h  -f  ac  +  ad. 

a^h  —  c  4-  d^        by  a. 

Ans. 

a^h  —  ac  -f  ad. 

a%  -{-  c  —  d        by  a. 

Ans. 

a^h  +  ac  —  ad. 

—  d'h-^c  +  d  by  a. 

Ans. 

—  a^'h  -f  ac  +  ad. 

—  a'^h  —  c  -\-  d  by  a. 

Ans. 

—  a^h  —  ac  -f  ad. 

—  a^h  —  c  —  d  by  a. 

Ans. 

—  a^h  —  ac  —  ad. 

—  d'h  —  c  —  d  by — 

a. 

Ans. 

a^h  -\-  ac  -\-  ad. 

—  a^h  —  c-\-d  by  — 

a. 

Ans. 

a^h  -\-  ac  —  ad. 

—  a^b-\-c-\-d  by  — 

a. 

Ans. 

a^h  —  ac  —  ad. 

a^b  +  c-j-  d        by  — 

a. 

Ans. 

—  aNj  —  ac  — •  ad. 

m^n  +  a  -f  2c    by  s. 

Ans. 

m"ns  -f-  as  -j-  2c,9. 

m^n  +  a  +  2c    by  3c 

Ans. 

Scm'n  -f  Sac  -f  Gc^ 

m^n  -f  a  +  2c    by  4wn. 

Ans. 

4??i^;i'^  +  4amn  -f  Smnc. 

8m  — 6a  +  llcby  2c;= 

Ans. 

16mc''  — 12ac='-f  220". 

MULTIPLICATION.  29 


CASE  III. 

42.  When  tlie  multiplicand  and  multiplier  are  both  compound  quan- 
tities. 

The  multiplicand,  as  in  the  two  preceding  cases,  is  to  be  repeated  as 
many  times  as  there  are  units  in  the  multiplier.  The  multiplicand  and 
multiplier  will,  in  genei*al,  be  made  up  of  some  positive  and  some  ncg:'- 
tive  terms.  Let  a  denote  the  sum  of  all  the  positive  terms  in  the  mul- 
tiplicand, and  b  the  sum  of  all  the  negative  terms.  Let  c  denote  the 
sum  of  the  positive  terms  in  the  multiplier,  and  d  the  sum  of  all  th(> 
negative  terms.  We  write  the  multiplier  beneath  the  multiplicand 
and  multiply  all  the  terms  of  the  one  by  all  the  terms  of  the  other. 

Thus, 


a 
c 

—  b 

—  d 

ac 

—  be 

—  ad  +  bd 

ac  —  be  —  ad  -f-  bd. 

To  explain  this  result,  let  us  drop  fur  a  moment  the  consideration  of 

—  d,  then  a  —  b  must  be  repeated  e  times.  From  what  has  been 
shown,  the  result  of  this  multiplication  will  be  ac  —  be.  But  we  were 
not  required  to  multiply  a  —  i  by  c  alone,  but  by  c  after  it  had  been 
diminished  by  d ;  hence,  the  result,  ac  —  be,  is  too  great  by  a  —  b 
taken  d  times.  To  correct  the  error,  then,  we  must  subtract  the  pro- 
duet  of  a  —  bhj  d  from  the  first  product  ac  —  be.  The  product  of 
a  —  b  by  -f  d,  from  what  has  been  shown,  will  be  +  ad  —  bd,  and 
to  indicate  that  this  must  be  subtracted,  we  write  it  in  parenthesis  with 
the  minus  sign  before  it.     The  whole  result  will  be  oc  —  be  —  (-f  ad 

—  bd)  =  ac  —  be  —  ad  +  bd,  since  the  signs  of  the  quantities  sub- 
tracted must  be  changed.  By  examining  the  result,  we  will  observe,  as 
before,  that  the  product  of  quantities  affected  with  like  signs  is  positive, 
and  that  of  quantities  affected  with  unlike  signs  is  minus. 

43.  It  is  found  most  convenient  to  arrange  both  polynomials  with 
reference  to  the  highest  or  lowest  exponent  of  the  same  letter.  Thus, 
x^  -{-  x^  +  X  -\-  a  is  arranged  with  reference  to  the  highest  exponent 
of  X ;  and  a  +  x  +  x^  +  x^  is  arranged  with  reference  to  the  lowest 
exponent  of  the  same  letter.  In  these  expressions,  x  is  supposed  to 
enter  to  the  zero  power  in  the  term  a,  as  will  be  explained  under  the 
head  of  Division. 
3* 


MULTIPLICATION. 


RULE. 


44.  Arrange  the  miiUipUcand  and  multiplier  tvith  reference  to  the 
highest  or  loicest  exponent  of  the  same  letter  (if  tliey  have  a  common 
letter'),  and  then  multiply  each  te^m  of  the  one  polynomial  by  each  term 
of  the  other  polynomial,  leg  inning  on  the  left.  Set  down  the  result  of 
the  midtiplication  of  the  second  term  of  the  multiplier  under  that  of  the 
first  term,  only  removed  one  place  further  to  the  right,  and  the  result 
of  the  multiplication  of  the  third  term  of  the  multiplier  %inder  that  of 
the  second,  and  contimie  the  operation  until  the  multiplication  is  com- 
plete.     Then  reduce  the  lohole  result  to  the  simplest  form. 


1.  X  +  a 


X?  +  ax 
—  ax  —  a? 


■:i?  +  0  —  d'-  =■  x^ — a^. 

.    X^  +    O.T^ X 

x^  -\-  ax  -\-  \ 

xb  +  ax'^  —  x^ 

+  ax'^  +  ax^  —  ax^ 

+  a;'  +   ax^  —  x 

cc5  +  2ax'^  +  ax^  +  o  —  x 


2.  x^  -{-  xy  -\-  a 
X  +y 

x^  +  x'y  +  ax 
-\-  x^y  +  ar/+  ay 


X*  +  Ix^y  +  ax  +  xy^  +  ay. 

a?  +  2xy  +  f 
X — y 

x'  +  2x^y  +  y'^x 
—  x^y  —  2y'^x  —  y^ 

*'  +  x'y  —  y'^x    —  / 


45.  It  will  be  observed  in  the  above  examples,  that  by  arranging 
with  reference  to  a  certain  letter,  and  by  removing  each  result  one 
place  further  to  the  right  than  the  preceding,  the  terms  which  reduce 
fall  immediately  the  one  below  the  other.  And  it  is  to  save  the  trouble 
of  looking  for  the  terms  which  reduce,  that  this  arrangement  and  sys- 
tem are  made.  It  will  also  be  noticed  that  in  every  product  there  are 
two  terms  which  do  not  reduce  with  other  terms,  viz.,  those  which 
result  from  the  multiplication  of  the  quantities  affected  with  the  highest 
and  lowest  exponents  of  the  arranged  letter. 

In  the  first  example,  x^  and  —  a^  are  the  irreducible  terms,  in  the 
second,  x?  and  ay. 


MULTIPLICATIOX.  31 


EXAMPLES. 

6.  Multiply  X  -\-  7/hj  X  -{-  1/. 

Ans.  x'  +  2xi/+7/\ 

7.  Multiply  X  +  y  by  cc  —  i/. 

Ans.  7?  —  1)^. 

8.  Multiply  x^  +  Ixy  +  ^^^  by  +  cc  +  y. 

.4ns.  x^  +  Zx^ij  +  ^n'^x  +  y^. 

9.  Multiply  X'  —  2xy  -\-  y^\>^  x  +  y. 

sins.  X?  —  x-y  —  xy'^  -\-  y^. 

10.  Multiplyx^  — T/^byx^+y. 

Ans.  x^  +  y^x"  —  x'y'  —  /. 

11.  Multiply  a-  +  lax  +  a;^  by  x  -^r  a. 

Ans.  a;'  +  Sux'^  +  3alt+  a'. 

12.  Multiply  .t'  +  a^"  +  3aa;  +  3«.7;2  by  x  +  a. 

.'1/)S.  x"  +  4ax'  +  6al<;2  +  \c{'x  +  a^ 

13.  Multiply  7x-^  +  2«  by  %y'-- 

Ans.  2.\oh/  +  6^7/^. 


14.  Multiply  4a;*  +  y^  by  4.c*  —  y*. 


.4«s.  16x^— /. 


15.  Multiply  x^  H-  o.c^  +  a^x  +  a'  by  2r/  +  2.f. 

^ns.  2a;''  +  hix^  +  4(/^.r  +  4a='a:  +  2^". 

16.  Multiply  x^  +  xy  +  /  by  1x  +  7y. 

^«s.   7.<'  +  14a-y  +  14a-y^  4-  ly^. 

17.  Multiply  a;'  +  ax'  +  a^x^  +  a'x  +  a<  by  4a;  +  4a. 

ylns.  4x*  +  Sax*  +  Sa^x'  +  8aV  +  Sa^x  +  4al 

18.  Multiply  2x'  +  2ax  +  2a'  by  3x  +  3y. 

Ans.   ^x^  +  Gax'  +  Gx^y  +  Ga'x  +  G«xy  +  Ga^- 

19.  Multiply  2a2  —  2^^  by  2a  +  26. 

.4«s.  4a3  +  4a^6  —  \aV  —  \h\ 

20.  Multiply  2x  +  3a  +  46  by  y  +  z. 

Ans.  2xy  +  Zay  +  46y  +  2x3  +  3a2  +  Ahz. 

21.  Multiply  2x  +  3a  +  46  by  2x  —  2y. 

Ans.  4x'  —  4.ry  +  6ax  +  86x  —  Qay  —  8iy. 

22.  Multiply  4/  +  4x^  +  4m?i  by  2x  +  2y  +  2z. 

Ans.  8y='  +  Syx'  +  Sywin  +  8x'  +  Sxy'  +  ^xmn  +  Sy''^  +  Sx's 
+  8m?tz. 


MULTIPLICATION. 


RemarJis. 


46.  It  -will  be  seen,  by  inspecting  tlie  above  results,  that  when  the 
two  polynomials  are  both  homogeneous,  the  product  is  also  homoge- 
neous, and  of  a  degree  equal  to  the  sum  of  the  dimensions  of  the  mul- 
tiplicand and  multiplier.  Thus,  in  Example  7th,  the  first  polynomial 
is  homogeneous,  and  of  the  first  degree ;  and  the  second  is  also  homo- 
geneous of  the  first  degree,  the  product  is  homogeneous  of  the  second 
degree.  In  Example  8th,  the  multiplier  is  homogeneous  of  the  second 
degree,  and  the  multiplier  is  homogeneous  of  the  first  degree;  the  product 
is  homogeneous  of  the  third  degree.  The  same  may  be  noticed  in 
several  other  examples. 

2d.  We  notice  also  that  when  the  coefl&cients  are  the  same  in  each 
term  of  the  multiplicand,  and  the  same  in  each  term  of  the  multiplier, 
and  all  the  terms  of  both  polynomials  are  positive,  that  the  sum  of  the 
coefiicients  in  the  product  will  be  equal  to  the  product  arising  from  mul- 
tiplying the  sum  of  the  coefiScients  in  one  polynomial  by  the  sum  of  the 
coefiicients  in  the  other.  Thus,  in  Example  16,  each  coefficient  of  the 
multiplicand  is  1.  and  each  coefficient  of  the  multiplier  7.  The  sum  of 
•the  coefficients  of  the  multiplicand  is  3,  and  that  of  the  multiplier  14 : 
the  sum  of  the  coefficients  in  the  product  is  3  X  14  =  42.  The  same 
law  may  be  noticed  in  Examples  15,  18,  and  22. 

47.  Examples  10  and  19  show. that  when  the  multiplicand  is  com- 
posed of  the  diff"erence  of  two  terms,  whose  coefficients  are  equal,  the 
algebraic  sum  of  the  coefficients  in  the  product  is  zero.  Examples  7 
and  21  show  that  this  sum  is  also  zero  when  the  multiplier  is  composed 
of  two  terms  with  contrary  signs  and  equal  coefficients. 

3d.  It  has  already  been  remarked  that  there  are  always  two  terms, 
which  do  not  reduce  with  any  other  terms.  We  can  only  reduce  similar 
terms,  and  when  the  two  polynomials  have  been  arranged  with  respect 
to  a  certain  letter,  the  products  of  the  extreme  terms  are  dissimilar  to 
the  other  partial  products.  The  whole  process  of  division  of  polynomials 
is  based  upon  this  fact,  and  it  ought  to  be  remembered.  By  attending 
to  the  above  laws  in  regard  to  the  product,  we  can  often  by  a  simple 
inspection  detect  errors  in  the  multiplication. 

There  are  three  theorems  of  great  importance,  which  must  be  com- 
mitted to  memory. 


MULTIPLICATION.  33 


Theorem  I. 


48.  The  square  of  the  sum  of  two  quantities  is  equal  to  the  square 
of  the  first,  plus  the  double  product  of  the  first  by  the  second,  plus  the 
square  of  the  second. 

Let  X  denote  the  firi<t  <[uanlity,  and  a  the  second,  then  .r  +  a  =  sum 
and  {x  +  «)^=(x  4-  a)  (.c  +  a),  which,  by  performing  the  multiplica- 
tion, will  be  found  equal  to  x^  +  lax  +  a^. 

Hence  (x  +  iif^=x^  +  lax  +  a^,  as  enunciated. 

So  (10  +  5/  =  10^  +  20  X  5  +  5^  =  100  +  100  +  25  =  225. 

And  (40  +  6/  =  40^  +  80  X  6  +  G^  =  1600  +  480  +  36  =  2116. 

In  like  manner  (x""  +  a"/  =  x^""  +  2x'"a'"  +  a^"". 

The  rule  may  be  extended  to  binomials  of  any  form. 

Thus,  (3a  +  hj  =  (3a)^  +  6a/j  -\-  V  =  Oa^  +  6«&  +  i'- 

(7x  +  ^yf  =  (7x)=+  14x  ibij)  +  {hyf  =  49x^  +  lOxy  +  25f. 

A  polynomial  may  be  squared  by  the  same  formula. 

Let  it  be  required  to  square  x  +  a  +  i/. 

Make  X  +  a  =  z. 

Then  X  +  a  +  jy  =  r:+7/  and  (x  -f  a  +  yf  =  (^  +  yf  =  :?  +  2.ry 

+r- 

Now  substitute  for  z  its  value  x  -\-  a,  and  we  have  (x  +  a  +  yf  = 
(^  +af+2  (x  +a)y  +  f  =  x'+  2ax  +  x' +  2x>/  +  2ay  +  //. 

Required  the  square  of  2^  -f  x  +  «t  +  n. 

Let  2}/  -{-  x  =  z,  and  m  +  '*  =  s. 

Then  (2y  +  x  +  m  +  nf  =  (z  +  sf  =  z' +  2zs  +  s' =  (2y  +  xf 
+  2  (2>/  +  X)  (m  +  n)  +  (m  +  nf  ==  4/  +  4yx  +  x'  +  4>/m  +  iyn 
-f  2xm  +  2.rn  +  ni^  +  2»i/(  +  n\ 

Required  the  square  of  oy  +  4x  +  m  +  2n-f  5. 

Represent  the  first  three  terms  by  a  single  letter,  and  the  last  two 
also  by  a  single  letter,  and  proceed  as  before. 

TlIEOHEM    II. 

49.  The  square  of  the  diiTorcnce  of  two  quantities  is  equal  to  the 
square  of  the  first,  minus  the  double  product  of  the  first  by  the  second, 
plus  the  square  of  the  second. 

Let  X  and  a  denote  the  quantities,  then  x  —  o  =  difference. 
And  (x  —  a)^  =  (x  —  a)  (x  —  a)  =  x^  —  2ax  +  a^,  as  enunciated, 
So(10  — 5)^  =  10^— 20x  5+  (5)2=^100  —  100  +  25  =  25. 
And(40  — 6)^  =  (40)^  — 80  X  6  +  6^  =  1600—480  +  36  =  1156. 


34     .  MULTIPLICATION. 

In  like  manner  (a;"  —  a"")^  =  x^'"  —  2x"'(r  +  a^'". 
llequircd  tlie  square  of  2x  —  a.  Ans.  4x'^  —  4ax  +  a^. 

Required  the  square  of  3a;  +  h  —  a. 

Ans.  (3x  +  hy  —  2  (Sx  +  h)a+  a^  =  dx^  +  Q>hx  +  1/—  Qax  —  2ah 
+  aK 
llequircd  square  of  a;^  —  a^.  Ans.  x^  —  2x^a?  +  a^. 


Theorem  III. 

50.  The  sum  of  two  quantities  multiplied  by  the  difference  of  the 
same  quantities,  is  equal  to  the  difference  of  their  squares. 

Let  X  and  a  denote  their  quantities. 

Then  x  -{•  a=  sum,  and  x  —  a  =  difference,  and  (.«  +  o)  {x  —  a) 
==  x^  —  a^,  by  performing  the  multiplication  indicated. 
So  (10  +  5)  (10  —  5)  =  (10/  —  (5)^  =  100  —  25  ==  75. 
And  (40  +  6)  (40  —  6)  =  (40)''  —  (6)*  =  1600  —  36  =  1564. 
Multiply  (4«  +  6)  by  (4a  — 6).  Ans.  16a^  —  6^ 

Multiply  la  +  /j  +  f  by  7a  +  6  —  c. 

Ans.  (7a  +  ly  —  c'  =  49a2  +  Uab  +  J/  —  c\ 
Multiply  x  -\-  y  +  z  +  m\)j  .K  -\-  y  +  z  —  m. 

Ans.  (,x  +1J  +  zj  —  m''  =  x'  +  f+  z"  +  2xy  +  2xz  +  2ijz  —  m". 
Multiply  X  -\-  y  ^r  z  -{-  m\)^  X  ■\-  y  —  z  —  m. 

Ans.  {x  +  yf  —  {z  +  mj-  =  a-^  -f  2xy  -\-  y'^  —  ^^  —  2mz  —  m'- 
■V 

Remarks. 

51.  If  we  omit  the  exponents  of  the  extreme  terms  in  the  expres- 
sion, x^  +  2«x  +  a^,  and  connect  these  terms  with  the  exponents  so 
omitted,  by  the  sign  of  the  middle  term  2oa-,  we  have  a;  +  a,  the  bino- 
mial, which  squared  gave  x?  -f  2aa;  +  a^-  In  like  manner,  if  we  omit 
the  exponents  of  x^  and  o^  in  the  expression,  a?  —  2aa;  -f  o?,  and  con- 
nect them  with  their  exponents  omitted  by  the  sign  of  the  middle  term, 
we  have  x  —  a,  the  binomial,  which  squared  gave  a;**  —  2ax  +  a;l 

In  like  manner,  if  we  have  given  the  difference  of  two  squares,  we 
can  readily  determine  the  quantities  which,  by  being  multiplied  together, 
gave  this  difference. 

Thus  n^  —  ?j^  =  (m  +  n)  (m  —  ?i),  and  r-  —  .s^-:=  (r  +  s)  (r  —  s). 

Omit  the  exponents,  connect  the  terms  by  the  sign  plus,  and  we  have 
the  sum  of  the  two  quantities ;  omit  and  connect  by  the  .sign  minus,  and 


FACTORING    POLYNOMIALS. 

WO  have  the  difference.  And  the  product  of  the  sum  by  the  difference, 
is  the  difference  of  the  squares.  These  remai-ks  are  preliminary  to  an 
important  subject. 

FACTORING   POLYXOMIALS. 

52.  It  is  often  a  matter  of  great  importance  to  resolve  a  polynomial 
into  its  factors.  The  reduction  of  expressions  can  often  be  effected  in 
this  way,  and  in  no  other.  Practice  alone  can  make  the  student  expert 
in  the  resolution  of  an  algebraic  expression  into  its  factors.  A  few  rules 
may,  however,  assist  the  beginner. 

53.  1st.  Look  out  for  the  terms  which  have  a  commoa  factor,  and 
write  them-  in  parenthesis,  as  a  multiplier  of  that  factor  j  next  look  for 
another  set  of  terms  also  having  a  common  factor,  and  write  them  in 
like  manner;  proceed  thus  until  all  the  terms  are  taken  up.  Observe 
whether  the  parenthetical  expressions  are  the  same,  if  so,  multiply  the 
common  parenthesis  into  the  ahjchraic  sum  of  the  terms  to  which  it 
serves  as  a  coefficient. 


1.  hx  +  ha->r  CX  +  ca  =  h  (./-  +  a)  +  c  (.c  +  o)  =  (x  +  a)  {h  +  c). 

2.  hx  +  ha  —  cx  —  ca  =  h{x -\- a)  +  (x^  a)  (— c)  =  (x  +  a)  {h 
-c). 

3.  —  hx  —  ha  -f  ex  +  ca  =  (—  h)  (.r  +  a)  +  (.c  +  a)  c  =  (x  +  a)  (r 

4.  hx  +  ex  -\-  ha  +  dx  +  da  -f  ca  =  (.c  -{.  a)h  -{■  (x  -\-  a)  c-\-  (^x-'r  a) 
d  =  (x+a)  (h+c-ird). 

5.  hx-\-  ex  -\-  ha  —  dx  —  d(t  -f-  ca  =  (x  +  o)  (h  -\-  c  —  d). 

6.  a^  +ax  +  hx-{-  ah  =  {x  +  a)  x  +(x-  -{-a')h  =  (x+  a)  (x  +  6). 

7.  x^  —  ax  -{■  hx  —  ah  =  (x  —  «)  (x  +  />). 

8.  x^  —  ax  +  hx  —  ah  +  x^  —  ax'  =  (x  —  a)(x+h  +  x"). 

9.  x^  —  ax  +  hx  —  ah  -f  mx^ — ajnx^  =  (x  —  o)  (x  +  h  +  mx^). 
10.  x^  —  ax  +  ix  —  cd)  +  amx^  —  mx^  =  (x  —  a)(x  -\-  h  —  mx^). 

54.  2d.  After  having  found  a  common  factor  or  parenthesis,  see 
whether  that  factor  may  not  admit  of  farther  reduction.  (Articles  48, 
49,  and  50.) 


3G  F  A  C  T  U  11 1  N  (i     P  U  L  Y  N  tJ  M  I  A  J.  S  , 


EXAMPLES. 


1.  W  +  Ihn  +  6  +  cn^  +  2c??.  +  c  =  h  (n^  +  2w  +  1)  +  c  (u'  +  2« 
+  !)  =  (]>  +  (■)  («2  +  2u  +  1)  =  (?,  +  c)  (n  +  1)1     (Article  48.) 

2.  bn''  —  2/';h  +  Z>  +  C7i'  —  2ai  +  c  =  {b  +  c)  («  —  1)''.   (Art.  49). 

3.  am"  +  hm^  —  an'  —  hit^  =  (a  +  h)  (vi' —ir)  =  (a  +  b)  (m  + 
7i)(^m  —  n).     (xVrticle  50). 

4.  am^  —  bm^  —  an^  +  bn^  =  (a  —  b')  (vi  +  n)  (m  —  «). 

5.  «=>  +  2n'  +  7i  =  n  (n  +  \)\ 

6.  n^  -f  2;i^  =  n^  (ji  +  2).     Admits  of  no  lower  reduction. 

7.  «='—  2n:2  +  7i  =  n  (n  —  If. 

8.  ii^  —  2n^  +  n  +  07111^  — 2mn  +  m  =  (n  +  m)  {ii  — 1)1 

9.  n^  —  271^  -\-  n  —  wi/t^  +  2mn  —  m  =  (ii  —  vi)  (n  —  1)^. 

10.  —  n^  +  2?t^  —  n  +  mii'  —  2mn  +  m  =  (m  —  n)  (n  — 1)1 

11.  _  „3  +  2»2—  ?i  —  m/i"  +  2??j?i  —  wi  =  —  (/I  +  m)  (?i  —  1)1 

12.  m^  —  n'  —  2c7i  —  c^  =  m^  —  (h  +  cf. 

We  have  now  the  differences  of  two  squares,  and  to  apply  the  formula 

x"  —  a^  =  (x  +  a)  (x-  —  a).      (Art.  50). 

We  see  that  m^  =  .r^,  or  on  =  .r,  and  (n  +  cf  =  ry^,  or  w  +  c  =  a. 
Hence  (x  -\-  a)  =  {m  +  ?i  +  c),  and  (.r  —  ff)  =  hn  —  n  —  c). 
Therefore  m^  —  (?i  +  cf  =  (vu  +  n  +  c)  (?n  —  n  —  c). 

13.  m^  — 11^  +  2cn.  —  c^  =  Hi^  —  (n  —  cf  =  (m  +  n  —  c)  (in  +  c 
—  ??).      In  this  Example  m  =  x,  and  n  —  c  =  a. 

14.  m-  +  2Z;??i  +  b^—  n""  —  2cn  ^  n"  =  (m  +  bf  —  (n  +  cf  =  (m 
+  b  +  n  -\-  c)  X  (m  +  b  —  n  —  c). 

55.  3d.  Expressions  may  sometimes  be  thrown  into  factors  by  an 
artifice,  when  they  do  not  at  first  glance  appear  to  admit  of  resolution 
into  factors.  One  of  the  simplest  contrivances  to  effect  a  decomposition 
into  factors  is  the  addition  to,  and  subtraction  of  the  same  quantity 
from  the  given  expression,  which  operation  docs  not,  of  course,  alter  its 
value. 

EXAMPLES. 

1.  x^^2x  —  l'i  =  x-  —  2x  +  1  — 15— l  =  a;^  — 2a:  -r  1  — 16 

=  (.r  — 1)^  — lG  =  (a'  — 1)^  — (4)=^^(x  — 1  +  4)    (a;— 1  —  4). 
Art.  50.     ={x  +  o)  (x.  —  b). 


FACTORING    POLYNOMIALS.  37 

In  this  example,  the  decomposition  was  effected  by  adding  +  1,  and 
subtracting  the  same  from  the  given  expression.  The  first  three  terms 
of  tlie  equivalent  expression  were  thus  made  a  perfect  square  =  (x  —  1^), 
and  the  whole  was  made  the  difference  between  two  squares. 

2.  x"  +  -ix  —  12  =  x'  +  -ix  +  4  —  12  —  ,4  =  (x  +  2)^  —  (4)^  = 
(x  +  0)  (X  —  2). 

3.  x'  +  2x  —  S  =x'  +  2x  +  l  —  S  —  l=  (x  +  lf  —  (Sy  = 
(.,  +  4)  (x  —  2). 

4.  x'  +  4x  —  21  =  x'  +  4x  +  4  —  21  —  4  =  (.f  +  2)^  —  (5/  = 

(x  +  7)  (x  —  n). 

5.  x^  +  8x  —  48  =  x'  +  Sx  +  IG  —  48  —  IG  =  (x  +  4/  —  (8)^ 
=  (x  +  12)(x  — 4). 

6.  x'  ^  10.r  +  24  =  x"  4-  10a:  +  25  +  24  —  2-5  =  (x  +  5)^  —  (1/ 
=  (x  -h  6)  (X  +  4). 

7.  ;y2  _u  8x  +  12  =  X'  +  8x  +1G  -f  12  —  IG  =  (x  +  4)=  —  (2/ 
=  (x  +  G)  (X  +  2). 

8.  x^  +  20x  +  84  =  x^  +  20x  + 100  +  84  —  100  =  (x  + 10/  — 
(4y  =  (x  +  14)  (re  +  G).  

9.  x'  4-  12x  +  2  =  x'  +  12x  +  30  +  2  —  3G  =  (x  +  G)^  —  V  (34/ 
=  (x  +  G  +  V34)  (x  +  6—  v/34).     ' 

10.  x'  +  3x  +  5=  x'  +  3x  4-  »  +  5—  9  =  C.r  4.  |)2  _  ^"(=^17/ 

=  (-*^  + 1  +  V— y)  (-r  +  5  —  v/— '.,')• 

11.  x^  —  lOx  4-  24  =  x^  —  lOx  4-  25  4-  24  —  25  =  (.r  _5/  — (1/ 

=  (,;_4)(x-6). 

12.  x^  —  8x  4-  12  =  x^  —  8x  +  ] G  +  1 2  —  IG  =  (x  —  4)^  —  (2/ 
=  (,c_2)(x-G). 

13.  x^  —  1  Ox  —  24  =  x'  —  lOx  +  25  —  24  —  25  =  (x  —  5)-  — 
(7/  =  (x  +  2)(x— 12). 

14.  x'  —  10./-  —  24  =  ./•'  —  10./;2  4-  25  —  24  —  25  =  ^x^  —  of  — 
(7/  =  (.r^4-2)0-^  — 12). 

15.  ^>-6  _j.  4,.3_  12  =  .,/•  +  4.f3  4-  4  —  12  —  4  =  (.,;'  4-  2)^  —  (4)" 
=  (x'4-G)(.f^  — 2). 

IG.  (f-  4-  2aIj  —  SI/  =  >r  +  -lab  -\-  Ir  —  W  =  (a  -f  If  —  (2hf  = 
(a  +  3i)  (a  —  h-). 

17.  a^  4-  2((h  —  m^  —  Ihm  =  rt^  4-  2ah  +  //  —  m^  —  2hm  —  1"  = 
{a  +  ly  —  (m  +  h'f  =  (a  4-  hi  +  2Z/)  (a  —  m). 

18.  a'  +  2ab  —  TO^  +  2bm  =  a"  +  2ab  4-  &'  —  m^  4  2bm  —  b^  = 
(a  +  bf—(in  —  bf  =^{a  +  m){a  —  m-\-  2b'). 

4 


D  I  ^'  I  S  I  0  N  . 


DIVISION. 

56.  Division  consists  iu  finding  how  many  times  one  quantity  is 
contained  in  another. 

The  quantity  divided  is  called  the  dividend,  that  by  which  it  is 
divided,  the  divisor,  and  the  result  obtained  the  quotient. 

57.  It  follows  from  this  that  the  quotient  multiplied  by  the  divisor 
must  give  the  dividend. 

The  quotient  is  said  to  be  exact  when  the  dividend  contains  the 
divisor  an  exact  number  of  times.  When  this  is  not  so,  the  quotient  is 
called  imperfect. 

It  will  be  seen  that  the  object  of  division  is  to  find  a  quantity  called 
the  quotient,  which,  when  multiplied  by  the  divisor,  will  give  the 
dividend. 

The  result  of  the  division  of  4rtx  by  2a  is  plainly  2x,  because  2a  X 
2a;  =  4aa",  the  dividend. 

So,      ■ — -  =  la,  because  7a  X  ac  =  7a^<:. 
ac 

But  — ; — —  gives  — 1x  for  a  quotient,  because  2a  X  ( —  2x)  =  —  4ax. 

i     ,  — 7a^c  ^      .  ^  „  2 

And =  — ia,  because  —  i a  x  ac  =  —  ia'^c. 

ac 

58.  And,  in  general,  since  the  quotient  into  the  divisor  must  give  the 
dividend ;  when  the  sign  of  the  quantity  to  be  divided  is  unlike  that  by 
which  it  is  divided,  the  result  will  be  negative,  and  when  the  sign  of 
the  quantity  to  be  divided  is  like  that  by  which  it  is  divided,  the  result 
will  have  the  positive  sign. 

Thus, J-  =  +  a,  because  +  a  X  —  h  =  —  ah. 

59.  In  Division,  then,  as  in  multiplication,  like  signs  produce  +, 
and  unlike  signs  — . 

There  are  three  cases  iu  Division. 


CASE  I. 

Whe7i  the  dividend  and  divisor  are  hoth  monomials. 
60.  Divide  a^  by  a  ;  we  are  to  find  a  quantity,  which,  multiplied  by 
a,   will   give  aj^.     This   quantity   is   plainly  a^,   because  a^  X  a  =  a^. 


DIVISION.  39 

And,  since,  in  multiplication,  we  add  the  exponents  of  like  letters,  Ave 
must,  in  division,  subtract  the  exponents  of  like  letters. 

Divide  6a'  by  2«,  the  result  is  obviously  3a^ ;  because  2a  X  oa^  = 
Qa".  And  since,  in  multiplication,  we  multiply  the  coefficients  together 
for  a  new  coefficient,  we  must,  in  division,  divide  the  coefficients  for  a 
new  coefficient. 

Divide  a'^b  by  a,  the  result  is  plainly  ah,  because  oh  X  a  =  ci^h. 
Divide  a^6V  by  a,  the  result  is  aiV.  And,  in  general,  if  there  arc 
letters  in  the  dividend  not  common  to  the  divisor,  they  will  enter  into 
the  quotient  with  their  primitive  exponents. 


RULE. 

61.  Divide  the  coefficient  «f  the  dividend  hi/  the  coefficient  of  the 
divisor  for  the  coefficient  of  the  quotient.  Write  after  this  neio  coeffi- 
cient all  the  letters  common  to  dividend  and  divisor  affected  icith  expo- 
nents equal  to  the  excels  (f  the  exponents  of  the  dividend  over  those  of 
the  same  letters  in  divisor,  and  icrite  the  letters  common  to  the  dividend 
only  with  their  primitive  exponents.  Give  the  quotient  the  sijn  -{■ , 
where  the  monomials  have  likf  siijns,  and  the  si(/n  — ,  ichcre  they  have 
unlike  signs. 

F.X.VMPLES. 

1.  Divide  x^y  by  x. 

2.  Divide  —  x^y  by  x. 
8.  Divide  —  .r'y  by  —  .r. 

4.  Divide  x^y  by  —  x. 

5.  Divide  4a"'Z>Pc  by  2tl 

6.  Divide  —  AOx^'yz^  by  —  20x>'"2-''. 

7.  Divide  a-?/  by  x'y. 

8.  Divide  X  y  by  a;,?/. 

9.  Divide  a;V  by  x'lf. 

10.  Divide  x\if  by  a^V- 

11.  Divide  .^s' by  2V. 

12.  Divide  z's^  by  zV. 

13.  Divide  lOOOa^Z-"  by  SOOa-^i-". 

14.  Divide  1000a-"Z;-"  by  —  SOOa-^i". 

1 

15.  Divide  a;  °  ?/"  by  .t^". 


Ans. 

x'y. 

Ans. 

-xV- 

Ans. 

^% 

Ans. 

-xhj. 

Ans. 

2a^U'c- 

Ans. 

2x-y 

Ans. 

x-y. 

Ans. 

X   ~y~\ 

Ans. 

x-^y-'. 

Ans. 

xy. 

Ans. 

z~^s. 

Ans. 

zs~\ 

Ans. 

2a'"'h'". 

Ans. 

—  2a~' 

40  DIVISION. 

02.  It  will  be  seen  that  the  division  of  monomials  is  impossible  when 
the  coefficient  of  the  dividend  is  not  divisible  by  that  of  the  divisor, 
and  when  the  divisor  contains  one  or  more  letters  than  the  dividend. 

63.  In  such  cases,  the  quotient  appears  in  the  form  of  a  fraction, 
which  may  admit  of  farther  reduction  by  striking  out  the  common  fac- 
tors.    Thus,  the  quotient  arising  from  the  division  of  7a^  by  'la^b,  is 

-r— rr,  because  — ^r-  X  la^b  is  plainly  equal  to  the  dividend  7«^.     But 
la^b  Za^b 

-r-^  may  be  reduced  to  -—  by  striking  out  the  common  factor,  a^. 
'lo?b       ''  'lb    ^  "^  ' 

In  like  manner,  la^b,  divided  by  2«//,  is  -—-  =   -— . 

64.  We  will  now  demonstrate  two  principles  which  will  enable  us 
to  reduce  such  expressions  still  lower. 

1st.  Any  quantity  raised  to  the  zero  power  is  equal  to  unity,  that  is, 

a°  =  1.     For  —  by  the  rule  for  the  exponents  in  division  is  equal  to 
ara-m  _  f,o^  j^j^^^  ^ny  quantity  divided  by  itself  is  also   equal   to  one. 

Hence,  —  =  a"  is  also  equal  to  1.     Therefore,  rt"  =  1. 

In  like  manner,  2"  =  1,  and  .  (1000)"  =  1,  and  (a  +  by  =  1,  &c. 

65.  2d.  A  quantity  may  be  transferred  from  the  numerator  to  the 
denominator,  or  from  the  denominator  to  the  numerator,  by  changing 

the  sign  of  its  exponent.      For  a~'°  may  be  multiplied  by  —  =  1, 

without  altering  its  value. 

„  a"         a"  1 

Hence,  a~  "■  =  a"""  x  —  =  —    =    — 

a™         «™         o™ 

But  we  know  that  «-">  =  —-—   which  we  have  seen,  is  also  eqiial  to  —  . 

1  ri™ 

The  quantity  a""  has  then  gone  from  the  numerator  to  the  denominator, 
by  changing  the  sign  of  its  exponent. 

In  like  manner,  a"*"""  may  be  multiplied  by  -3-  =  1,  without  altering 

its  value. 

Hence, 


So, 


o™ 

Cf-" 

a" 

1 

a™ 

T 

- 

=  a"" 

X 

a"" 

— 

a~" 

,  — 

rt~' 

1 

1 

z-p 

z-p 

z-f 

X 

~  p 

ZP 

zv 

3— p 

z" 

T 

DIVISION.  41 

The  quantity  z^  has  passed  from  the  denominator  to  the  numerator 
by  changing  the  sign  of  its  exponent. 

1  1  Z^  Z^  2P 

'  z~'^         z~''         z^         z°  1 

66.  By  the   first   principle,  the  quotient  in   Example   15  may  be 

_L-2  _L-2 

changed  into  x"      .  1  =  x-  ■> 

By  the  second  principle,  the  quotient  of  Icrb  by  2a^b  may  be 
changed  into  -;j^l>~\  and  the  quotient  of  la'^b  by  2«Z/^  may  be  changed 

into  —ah~\ 

Divide  x^t/  by  x^i/^c.  Ans.  a~'^~'c~',  or 

67.  In  this  example,  and  in  all  similar  examples,  when  the  divisor 
contains  a  letter  or  letters  not  contained  in  the  dividend,  we  may  con- 
ceive this  letter,  or  these  letters,  also  to  enter  into  dividend  raised  to 
the  zero  power,  and  to  execute  the  division,  we  have  only  to  subtract 
the  exponents  of  like  letters,  as  before.  Thus,  x^y  may  be  written 
x^i/c°,  and  we  have  only  to  subtract  1,  the  exponent  of  c-  in  the  divisor, 
from  o,  the  exponent  of  c  in  the  dividend.  The  result  will  be  o  —  1, 
or  — 1. 

Divide  x^y  by  a?b\* ;  then  x^y  =  x''ya''b°c'',  and  the  result  will  be 
x'ya-'b-'c-". 

68.  Strictly  speaking,  then,  there  is  but  one  case  in  which  the  divi- 
sion of  monomials  will  not  give  an  entire  quotient,  and  that  is,  when 
the  coefficient  of  the  dividend  is  not  exactly  divisible  by  the  coefficient 
of  the  divisor. 


CASE  II. 

69.  When  the  dividend  is  a  compound  quantity,  and  the  divisor  a 
monomial. 

The  dividend  may  be  regarded  as  the  product  of  each  term  of  the 
quotient  sought  by  the  monomial  divisor;  hence,  to  find  this  quotient, 
we  must  divide  each  term  of  the  dividend  by  the  divisor. 
4* 


42  DIVISION. 


70.  Divide  each  term  of  the  dioidcnd  separately  hy  each  term  of  the 
divisor,  as  in  Case  I.,  and  conned  the  partial  quotients  hi/  their  ap- 
propriate signs. 

EXAMPLES. 

1.  Divide  6x-/  —  ^xhjz  +  802  by  2.z. 

Ans.  ^xhfz-'  —  1x?ij^\a. 

2.  Divide  oc^  +  2a.c  +  a^  by  x.  Ans.  ic  +  2a  +  o^.»~'. 

3.  Divide  r?  +  .r"+'  +  a-''+^  +  x^"^  by  a:?. 

71h,s.   i  +  x  +  :r-  +  .7/ 

4.  Divide  x-^  +  x-p+'  +  rc-P+''  +  x-'^"^^  by  a;-p. 

^??.s.   1  +  .T  +  x^  +  x' 


5. 

Divide 

X- 

-P-! 

.T-P- 

-2_ 

-  X' 

-p-3 

.a:- 

p-4 

by 

.TP. 

yl 

is. 

.X-2P-'  — 

.T" 

-2r-2 

X" 

-2p-3 

6. 

Divide 

X~ 

-p-1 

— 

a^-P- 

-2_ 

-X' 

-;^3 

Ans 

p-4 
X 

by 

-1  _ 

ir-p. 

X 

7.  Divide  S.r''  +  2y  -f  2^  by  2;^.  Ans.  -^xh-'  +  yrr^  ^~ 

3  z" 

8.  Divide  3x'  +  2j/  +  .^'^  by  2^-'.  ^?2S.  —  xH  ^  yz  ■\-  — 

9.  Divide  \^a^l?  —  \^ahx  +  15o?>.Ty  by  5a6. 

^4?(s.  SaZ*  —  2.T  -(-  3a;?/. 
10.  Divide  A^a%  —  lOa^^x-  H-  15aZ>,ry  by  —  5a?;. 

Ans.  —  8a  +  2a;  —  3x7/. 


CASE  III. 

71.   When  the  dividend  and  divisor  are  both  polynomials. 

We  must  keep  in  view  that  the  dividend  is  the  collection,  after  addi- 
tion and  subtraction,  of  the  partial  products  arising-  from  multiplying  each 
term  of  the  quotient,  when  found,  by  each  term  of  the  divisor.  If,  then, 
we  can  find  a  true  term  of  the  quotient  and  multiply  it  into  each  term 
of  the  divisor,  we  will  form  so  much  of  the  dividend  as  was  composed 
of  the  product  of  this  term  by  the  whole  divisor.  And  when  we  have 
subtracted  this  product  from  the  dividend,  the  remainder  will  be  made 
up  of  the  partial  products  of  the  remaining  terms  of  the  quotient  not 
yet  found  by  each  term  of  the  divisor.     Now,  if  we  can  find  another 


DIVISION.  43 

true  term  of  the  quotient  and  multiply  it  into  the  divisor,  the  product 
will  be  so  much  of  the  dividend  as  is  made  up  by  the  multiplication  of 
this  second  term  of  the  quotient  by  each  term  of  the  divisor.  And  if 
■we  subtract  this  product  from  the  first  remainder,  the  new  remainder, 
if  there  be  any,  will  be  so  much  of  the  dividend  as  is  made  up  of  the 
product  of  the  remaining  terms  of  the  quotient  not  yet  found  by 
each  term  of  the  divisor.  We  can  thus  regard  each  remainder  in 
succession  as  a  new  dividend  until  all  the  terms  of  the  quotient  are 
found. 

72.  The  whole  difficulty  then  consi.^ts  in  finding,  in  succession,  true 
terras  of  the  quotient.  Now,  we  have  seen  that  in  the  product  of  poly- 
nomials there  are  always  two  terms  which  are  dissimilar  from  the  other 
terms,  and,  consequently,  irreducible  with  them.  They  arc  the  terms 
arising  from  the  multiplication  of  the  two  terms  in  the  multiplicand  and 
multiplier  affected  with  the  highest  and  lowest  exponent  of  the  same 
letter.  If,  then,  we  divide  that  term  of  the  dividend  which  contains 
the  highest  exponent  of  a  certain  letter  by  that  term  of  the  divisor 
which  contains  the  highest  exponent  of  the  same  letter,  we  are  sure  of 
getting  a  true  term  of  the  quotient.  For  this  terra  of  the  dividend  has 
been  formed  without  reduction  from  the  multiplication  of  the  corres- 
ponding term  in  the  divisor  by  the  quotient  found. 

73.  In  like  manner,  if  wc  divide  the  terms  afiected  with  the  lowest 
exponent  of  the  same  letter,  the  one  by  the  other,  we  are  sure  of  get- 
ting a  true  term  of  the  quotient  sought.  For  this  reason,  the  dividend 
and  divisor  are  arranged  with  reference  to  the  highest  or  lowest 
exponent  of  the  same  letter,  generally  with  reference  to  the  highest. 
AVhen  so  arranged,  the  first  term  of  each  successive  remainder  will  con- 
tain the  highest  or  lowest  exponent  of  the  letter  according  to  which  the 
polynomials  are  arranged,  and  will,  when  divided  by  the  fir^t  term  of 
the  divisor,  give  a  true  terui  of  the  quotient. 

The  divisor  in  Algebra  is  written  on  the  right  of  the  dividend,  and 
the  quotient  imraediately  under  the  divisor. 
Divide  a^  +  2ah  +  h^  hj  a  +  h. 

Dividend.        Divisor.  Dividpud.  Divisor. 

or  i^  -f  2ha  +  a-  \h  +  a 

h^  +     ha  \  h  -u  a 


2ab  +  h^ 

<d> 

a  +  h 
a  -f  h 

+ 

ah  -f  V' 
ah  +  h' 

1  Quotinit. 

ha  +  a^  \   Quotient. 
ha  +_a^ 
0     +  [)    Rrinainder.  0     -f  0  Remainder. 


44  DIVISION. 

74.  We  see,  that  by  arranging  dividend  and  divisor  with  reference 
to  the  ascending  or  descending  powers  of  the  same  letter,  and  dividing 
the  first  term  of  the  dividend  and  the  first  of  the  remainder  by  the  first 
term  of  the  divisor,  we  necessarily  get  true  terms  of  the  quotient.  But 
if  the  polynomials  are  not  arranged  we  will  not  get  true  terms  of  the 
quotient.     Write  the  dividend  and  divisor  thus, 

Dividend.  Divisor. 


V  +  h'  h 


a 


—    —  &c.,  Quotient. 


a 

We  get  —  for  the  first  term  of  the  quotient,  which  is  not  a  true 

result.     It  is  also  plain  that  the  division  would  never  end. 

75.  Since  each  successive  remainder  may  be  regarded  as  a  new  poly- 
nomial, to  be  divided  by  the  divisor,  we  can  change  the  arrangement 
of  the  remainders  at  pleasure,  provided  we  make  a  corresponding  change 
in  the  divisor.  lu  the  example  above  we  might  arrange  the  remainder 
ah  +  h"^  with  reference  to  h,  and  write  it  V^  +  ah,  provided  we  change 
the  arrangement  of  the  divisor,  and  write  it  6  +  a. 


^  RULE. 

76.  Arrange  the  dividend  and  divisor  with  reference  to  the  ascend- 
ing or  descending  powers  of  the  same  letter,  and  then  divide  the  first 
term  on  the  left  of  the  dividend  hy  the  first  term  on  the  left  of  the  divi- 
sor.     This  will  give  the  first  term  of  the  quotient. 

Multijyly  this  term  into  each  term  of  the  divisor,  and  suhtract  the 
product  from  the  dividend.  Divide  the  first  term  of  the  remainder  hy 
the  first  term  of  the  divisor  for  the  second  term  of  the  quotient.  Multi- 
ply this  term  into  the  divisor,  as  hefore,  and  suhtract  the  pjruduct  froin 
the  first  remainder. 

Continue  this  process,  dividing  the  first  term  of  each  remainder  hy 
the  first  term  of  the  divisor  until  we  get  a  remainder,  zero,  when  the 
division  is  said  to  he  exact. 

But,  if  the  exponents  of  the  letter,  according  to  which  the  arrange- 
ment is  made,  are  all  positive,  and  the  first  term  of  any  remainder  is 


DIVISION.  45 

not  divisible  hy  the  first  term  of  the  divisor,  the  quotient  is  incomplete  ; 
or,  -if  some  of  the  exjwnents  are  nerjative,  and  the  letter,  according  to 
which  the  arrangement  has  been  made,  has  disappeared  from  any  of 
the  successive  remainders,  the  quotient  is  also  incomplete. 


EXAMPLES. 

1.  Divide  x='  +  Sax^  +  Sa^x  +  a""  hy  x  +  a. 

Ans.  x'^  +  2ax  +  a*. 

2.  Divide  x*  +  4«j;'  +  Gr/^x^  +  Aa^x  +  x'  by  x  +  a. 

Ans.  a;»  +  3ax*  +  3«^x  +  a\ 

3.  Divide  x^  —  ^x^y  +  'ixy^  —  y  by  x  —  y. 

Ans.   x^  —  'l.ry  +  y'^. 

4.  Divide  4a==  —  I''  by  2a  +  h. 

Ans.   2a  —  b. 

5.  Divide  x"  —  y""  by  x  —  y. 


X   —y 
G.  Divide  x'  —  y^  by  x^  —  y^. 

Ans.  x«  +  xY  +  ./.y  +  /. 

7.  Divide  x"  —  y'^  by  x-  —  /. 

Ans.  x'"  +  x^^^  +  xy  +  xy  +  ^y  +  ^"'• 

8.  Divide  x'  —  y  by  x^  —  y\ 

Ans.  x^  -\-  xy  +  xy*  +  xy*  — ;/ 

9.  Divide  x'  —  ?/^  by  x'  —  y^. 

Ans.  X*  +  xy  4-  y^. 

10.  Divide  x^  +  /  by  x"  +  y\ 

Ans.  x«  —  xV^  +  xy  —  /  +  2^y^ 

•T^  +  f. 

11.  Divide  x'^'  +  y'^  by  x"  +  /. 

Ans.  x'"  ~  xy  +  xy  —  xy  +  xy  —  y'"  +  2//'^ 

12.  Divide  x'  +  y  by  x"  +  /. 

-4??s.  x^  —  xy  +  xy*  —  xy*  +  ?/ 
x^  +  /. 

13.  Divide  x®  +  y^  by  x'  +  y\ 

J.ns.  X*  —  x'^'  +  y^- 


46  DIVISION. 

14.  Divide  x''  +  if^  by  x"  -\-  f. 

Ans.  .t'"  —  x'l/  +  y"», 

15.  Divide  .r'^  +  y'^  by  x"^  +  if. 

Ans.  x'^  —  .ry  +  a;y  —  xV  +  y'\ 

16.  Divide  7ff.r  —  lai/  +  7«6  +  Z>x  —  hy  -\-  Ir  by  Ta  +  h. 

Ans.  X  —  y  -{-  h. 

17.  Divide  x'^y-^  +  a;  '  +  :^-y  +  ?/-'  by  a;^  +  if. 

Ans.  x-^  +  ?/-^. 

18.  Divide  14.x»  —  14/  by  Ix^  —  If. 

Ans.  2x«  +  2xy  +  2xy  +  2/. 

19.  Divide  a^  —  2nx  +  a;^  +  2f/y  —  Zxy  by  2y  +  a  —  x. 

Ans.  a  —  X. 

20.  Divide  ax^ — 2>arx-\-ar^+hx'^ — '2hi\x-\-hr^-\-cr'^ — '2crx-\-cx?-  hyx — r. 

Ans.   {x  —  r)  (a  +  i  +  c). 

21.  Divide  s'^  —  2sx  -[■  x^  -\-  rs  —  2xr  -\-  r^  -\-  rshj  s  —  x  -\-  r. 

Ans.  s  —  X  -\-  r. 

22.  Divide  50x«  +  50/  by  25^'  +  25/. 

Ans.  2x'  —  2^^/'  +  2f. 
2.3.  Divide  (a  -{- h -\- c)  (x  —  rf  by  (a  J^  h  -\- c)  {x  —  r). 

Ans.  X  —  r. 

24.  Divide  ^V'  +  ^'^  +  oc"  +  y  +  fh  +  x'-f  by  a;'  +  /. 

Ans.  y~^  -\-  Jj  -{■  x~'^. 

25.  Divide  A^x^  —  64^^^^  by  Ix^  +  8^y. 

^Hs.  7icy  —  8r^. 

26.  Divide  a«  —  h^  by  a^  +  V. 

Ans.  a'  —  oH;'  +  h'  —  2^ 

a'  +  //. 

27.  Divide  a'"  —  ?/°  by  a^  +  h\ 

Alls,   a'  —  a'l/  +  a'h*  —  a%'  +  h' —  2h^ 

a^  +  lA 

28.  Divide  a'  —  I'  by  n^  +  ?>'• 

Ans.   a"  —  a'Jr  +  a'b'  —  fA 

29.  Divide  a''  —  V  by  «='  +  h\ 

Ans.   a^  —  a'^I/  +  a'Z*«  —  6'. 

30.  Divide  a'°  —  l'°  by  a'  +  ?/. 

31.  Divide  4  by  1  +  :r. 

Ans.  4  —  4.r  +  4^2  —  4j-'  +  4a;'' 

^1  +  X. 


DIVISION.  47 

32.  Divide  4  by  x  +  1. 

Ans.  4./;-'  —  4.r-^  +  4.r-^  —  4.r-^ 


a;  +  1. 


33.  Divide  m  —  ?ix  +  px^  —  ^x^  by  1  +  a-.  * 

Ans.  m  —  (m  +  n)  x  +  (??^  +  n  -\-  p)  x^  —  otlier  terms. 

34.  Divide  1  +  x  by  1  —  :r. 

Ans.  1  +  2x  +  2.f2  +  2.6='  +  &c. 

35.  Divide  1  —  x  by  1  +  x. 

Ans.   1  —  2.r  +  2.v;^  —  2..='  +  &c. 
30.  Divide  x'=  +  «^  +  ax  +  1  +  xi/  by  x  +  o. 

Ans.  .r  -\-  y  -^  1 

X  -\-  a. 
37.  Divide  x^  +  x^  +  5  by  x='  +  2. 

J»S.    .T  +  1   +  3    —  2.T 

i?f?na?-/is. 

77.  I.  When  the  exponents  of  the  arranged  letter  in  dividend  and 
divisor  are  all  positive,  the  division  will  not  be  exact  if  the  first  terra 
of  any  of  the  reinaiudcrs  docs  not  contain  the  arranijed  letter  to  a  higher 
power  than  it  is  contained  in  the  divisor.  "When,  therefore,  we  got  a 
remainder  in  which  the  first  terra  contains  the  arranged  letter  to  a  lower 
degree  than  it  is  contained  in  the  divisor,  we  need  proceed  no  further, 
for  we  can  never  arrive  at  an  exact  quotient. 

If  some  of  the  exponents  of  the  letter,  according  to  which  the  poly- 
nomials are  arranged,  are  negative,  the  last  rule  will  not  hold  good,  for 
the  result  of  the  division  will  be  a  term  affected  with  a  negative  expo- 
nent, and  there  ought  to  be  one  or  more  such  terms  in  the  quotient, 
[f,  however,  the  letter,  according  to  which  the  ai'rangement  has  been 
made,  has  disappeared  from  any  of  the  successive  remainders,  we  need 
proceed  no  further. 

II.  Examples  6,  7,  and  9,  show  that  the  difference  between  the  like 

powers  of  two  quantities  is  divisible  by  the  like  powers  of  a  lower  degree 

of  those  quantities,  when  the  common  exponent  of  the  dividend,  divided 

by  the  common  exponent  of  the  divisor,  gives  an  exact  quotient.     And 

they  show  that  the  number  of  terms  in  the  quotient  is  expressed  by  the 

quotient  of  the  exponents.     Thus,  in  example  6  there  are  four  terms 

8 
in  the  quotient,  the  same  as—  the  quotient  of  the  exponents.     In  ex- 


48  DIVISION. 

ample  9,  the  quotient  of  the  exponents  is  three,  and  the  division  is 
exact,  with  three  terms  in  the  result. 

III.  Examples  13,  14,  and  15,  show  that  the  sum  of  the  like  powers 
of  two  quantities  is  divisible  by  the  sum  of  the  like  powers  of  the  same 
quantities  when  the  result  of  the  division  of  the  exponents  is  odd. 

IV.  Examples  28,  29,  and  30,  show  that  the  diiFerence  between  the 
like  powers  of  two  quantities  is  divisible  by  the  sum  of  their  like  powers 
when  the  quotient  of  the  exponents  is  even.  In  the  last  two  cases,  as 
in  the  first,  the  quotient  of  the  exponents  gives  the  nuuibor  of  terms  in 
the  result. 

V.  If  all  the  exponents  of  the  dividend  and  divisor  are  positive,  we 
can  tell,  by  a  simple  inspection,  in  many  examples,  whether  the  divi- 
sion is  possible;  when  the  extreme  terms  of  the  arranged  polynomials 
are  not  divisible  by  each  other,  the  quotient  will  not  be  exact.  Take 
^  -f  ^Ixy  +  'if',  to  be  divided  by  ^  -^  tf ;  the  division  is  impossible, 
because  the  exponents  of  the  result  must  be  positive ;  and  oi^  by  x*  will 
give  a;"'.  It,  of  course,  does  not  follow  that  the  division  will  give  a 
complete  quotient  when  the  extreme  terms  are  divisible  by  each  other. 
Take  a;^  -f  4  +  if,  to  be  divided  by  .-r  ■\-  y )  the  extreme  terms  are 
divisible  by  each  other,  but  the  quotient  of  the  polynomials  is  not 
exact. 

PRINCIPLES  IN  DIVISION. 

78.   I.  We  will  now  show  that  the  difference  between  the  like  powers 

of  two  quantities  will  be  divisible  by  the  difference  of  their  like  powers 

of  a  lower  degree,  whenever  the  quotient  of  the  common  exponent  of 

the  dividend,  by  the  common  exponent  of  the  divisor,  is  exact.     That 

is,  that  a"'  —  If"  is  divisible  by  a"  — ^  i",  when  m  is  exactly  divisible 

by  n. 

„m  _  ^^m  I  „"— ^;". 


a'"~"h"  —  b"'  =  h"  (a™  "  —  h"'   "). 

Performing  the  division,  we  get  a"'~"  for  a  quotient,  and  l>"  (a'"-"  — 
6"—°)  for  a  remainder.  If  this  remainder  is  divisible  by  a"  —  i",  the 
dividend  will  also  be  divisible  by  a"  —  l" ;  for,  represent  the  dividend 
by  D,  the  divisor  by  d,  the  quotient  by  q,  and  the  remainder  by  R. 
Then  a"  —  h"",  or  J)  =  qd  +  B,.  Now,  qd  is  plainly  divisible  by  d,  and 
if  R  be  also  divisible  by  d,  the  first  member  must  also  be  divisible  by 
d,  otherwise  we  would  have  the  sum  of  two  entire  quantities  equal  to  a 


DIVISION.  49 

fraction.  Kence,  in  general,  if  the  remainder  is  divisible  by  the  divi- 
sor, the  dividend  will  also  be  divisible  by  it.  Then,  if  we  can  prove 
that  the  remainder,  b"  (a"""  —  b'"~°),  can  be  divided  by  a"  —  Z>",  we 
can  prove  that  the  dividend  can  also  be  divided  by  a'  —  1°.  But  if  the 
factor,  a'"~°  —  i"""",  is  divisible  by  the  divisor,  h"  times  that  factor  will 
also  be  divisible  by  the  divisor ;  and  the  remainder  giving  an  exact  quo- 
tient, the  dividend  will  also  give  an  exact  quotient.  That  is,  a""  — J" 
will  be  divisible  by  a"  —  b"  whenever  a""""  —  i"""  is  divisible  by  a"  — 
b";  or,  in  other  words,  if  the  difference  of  the  like  power  of  two  quanti- 
ties is  divisible  by  the  difference  of  the  like  power  of  the  n*"*  degree, 
the  difference  of  the  like  powers  of  a  degree  higher  by  n  will  also  be 
divisible  by  the  difference  of  the  n**"  degree.  But  we  know  that  a"  — 
b"  is  divisible  by  itself,  a°  —  b" ;  hence,  by  the  principle  just  demon- 
strated, a^°  —  b^"  mu.st  be  divisible  by  a"  —  b".  And  since  a*  —  i*"  is 
divisible  by  a"  —  b",  a^"  —  i""  must  also  be  divisible  by  a"  —  b",  and 
so  on,  for  powers  of  a  degree  greater  by  n,  until  it  finally  reaches  and 
divides  a™  —  b'".  This  power  must  eventually  be  reached,  because  m 
is  supposed  to  be  a  multiple  of  n. 

79.  In  Example  8,  x'' — y,  divided  by  j:^  —  i/',  the  quotient  was 
not  exact,  because  m  or  7  was  not  a  multiple  of  n  or  2. 

In  Examples  6,  7,  and  9,  the  quotients  were  exact,  because  in  each 
case  m  was  a  multiple  of  n. 

80.  It  is  plain  thatj^a"" — pb""  can  be  divided  by  qa"  —  qb"  whenjj 
and  m  are  multiples  of  q  and  n.  For  we  can  put  the  expressions 
under  the  form  of  p  {a"'  —  b""),  and  q  (a"  —  i°),  and  the  quotient 
will,  of  course,  be  exact  when  j^  will  divide  q,  and  a"  —  b"'  will  divide 
a"  —  b\ 

Thus,  in  Example  18  we  found  Wx^  —  14//  divisible  by  Ix^  —  ly'^ : 
the  dividend  can  be  written,  14  (x^  —  y),  and  the  divisor  7  (.c^  —  y^, 
and  since  14  will  divide  7,  and  x^ — y*  will  divide  .'-^  —  ?/^,  the  quo- 
tient is  exact. 

81.  The  demonstration  holds  good  whenever  n,  added  to  itself  a  cer- 
tain number  of  times,  will  produce  m,  and  is  therefore  true  for  quanti- 
ties affected  with  negative  and  fractional  exponents. 

Divide  a"'  —  b-'  by  cr^  —  b-\  Ans.  a'^  4-  b-\ 

Divide  o-«  —  i-6  by  a-'  —  b-\  Ans.   a'^  +  i"^. 

Divide  «-«  —  b''  by  a~^  —  b'".  Avs.  a"*  +  a'^'b-^+b-^. 

5  D 


50  DIVISION 

Divide  a-  —  h^'  by  J  —  tl  Ans.  J  +  h\ 

Divide  a^  —  l^  by  a^  —  5  -  Ans.  a  +  a't^  +  b. 

Dividea®  — i'sby  a*  — 6*. 

^«s.  J  +  Jl^  +  Jb^  +  a%^  +  b^- 

82.  II.  The  difference  of  the  like  powers  of  two  quantities  is  divisi- 
Dle  by  the  sum  of  their  like  powers  of  a  lower  degree,  when  the  result 
of  the  division  of  the  common  exponent  of  the  dividend  by  the  common 
exponent  of  the  divisor  is  an  even  number;  which  is  evident  from  the 
principle  of  Article  50. 

83.  III.  The  sum  of  the  like  powers  of  two  quantities  is  divisible 
by  the  sum  of  their  like  powers  of  a  lower  degree,  when  the  quotient 
arising  from  dividing  the  common  exponent  of  dividend  by  the  common 
exponent  of  the  divisor  is  an  odd  number. 

That  is,  «"■  +  b""  is  divisible  by  a"  +  V",  when  m,  divided  by  rt,  is 
an  odd  number. 

Performing  the  division,  we  will  have  — Z*"  (a""""  —  J""")  for  a  re- 
mainder, and  whenever  the  factor,  a"""  —  i"*""",  is  divisible  by  a"  +  6", 
the  dividend  a™  +  b'^  will  be  divisible  by  the  same  divisor.  But  we 
have  just  shown  that  d"""  —  i"""  can  be  divided  by  a"  +  Z)°  when 

m  —  n   .  ,  in  —  n  .  m 

IS  an  even  number,  and  smce is  the  same  as 1, 

n  n  71 

„  'ni            ,        , ,  .               1             .          „  ???-  —  n  . 
the  quotient  of  —  must  be  odd  in  case  the  quotient  of is  even. 

Hence,  the  truth  of  the  proposition.     Thus,  in  Examples  13,  14,  and 

15,  the  quotients  were  exact,  because  for  each  of  these  examples  — 

was  an  odd  number.  So,  in  like  manner,  x^  +  a",  x^  +  a^,  x}  +  a', 
&c.,  are  divisible  by  a;  +  a,  and  give  3,  5,  7,  &c.,  terms  in  the  quo- 
tient, with  signs  alternately  plus  and  minus. 

84.  IV.  It  is  plain  that  the  sum  of  the  like  powers  of  two  quantities 
cannot  be  divided  by  the  difference  of  their  like  powers ;  that  is,  a"'  + 
i"  cannot  be  divided  by  a"  —  Z»",  because  it  is  not  possible  to  decom- 
pose the  sum  of  two  quantities  into  factors,  one  of  which  will  be  a  dif- 
ference between  the  quantities. 


AIC     FRACTIONS.  51 


ALGEBRAIC     FRACTIONS. 

85.  A  fraction  is  a  hrohcn  part  of  unity.  The  denominator  denotes 
the  number  of  equal  parts  into  which  the  unit  has  been  broken  or 
divided,  and  the  numerator  expresses  the  number  of  these  equal  parts 
taken.  Thus,  the  fraction  f  indicates  that  the  unit  has  been  divided 
into  three  equal  parts,  and  that  two  of  these  parts  have  been  taken.    In 

like  manner,  the  fraction  —  indicates  that  the  unit  has  been  divided 

b 

into  h  equal  parts,  and  that  a  of  these  parts  have  been  taken. 

8G.  Every  quantity  not  expressed  under  a  fractional  form  is  called 
an  entire  quantity. 

87.  An  expresssion,  made  up  in  part  of  an  entire  quantity,  and  in 
part  of  a  fraction,  is  called  a  mixed  quantity.     Thus,  4  +  J,  and  a  — 

—  are  mixed  quantities. 
c 

88.  A  proper  fraction  is  one  in  which  the  numerator  is  less  than  the 
denominator.  An  improper  fraction  is  one  in  which  the  numerator  is 
greater  than  the  denominator. 

89.  A  simple  fraction  is  one  whose  numerator  and  denominator  arc 

simple  quantities.     Thus,  —  is  a  simple  fraction. 

90.  A  compound  fraction  is  one  which  has  a  compound  expression 

in  the  numerator  or  denominator,   or  in  both.     Thus,  ,  , 

'       c     ^  c  -\-  d 

and  -j  are  compound  fractions. 

91.  The  minus  sign  before  a  fraction  changes  the  signs  of  all  the 

terms  in  the  numerator.     Thus, is  equivalent  to . 

b  b 

92.  A  few  of  the  principles  involved  in  operations  upon  fractions 
will  now  be  demonstrated. 

I.  The  multiplication  of  a  fraction  by  an  entire  quantity  is  effected 
by  multiplying  the  numerator,  or  dividing  the  denominator  by  the 
entire  quantity. 

For  to  multiply  —  by  r,  is  to  repeat  — ,  c  times,  and  since  —  taken 


02  ALGEBRAIC     FRACTIONS. 

twice,  is  V>  tlivce  times  is  — ,  &c,  the  result  of  multiplyin<2;  — ,  c  times 

oc 
will  plainly  ba  — .  The  multiplication  can  also  be  performed  by  divid- 
ing the  denominator  by  c ;  for,  if  we  divide  the  denominator  by  f,  and 
q  is  the  quotient  of  the  division,  then  the  unit  will  be  divided  into  c 
times  fewer  parts  than  before,  and,  of  course,  each  part  is  c  times  as 
great  as  before.  Now,  if  we  write  a  over  q,  we  will  have  taken  as 
many  parts  of  the  unit  as  before ;  and  since  each  part  is  c  times  greater, 

the  fraction  —  is  plainly  c  times  greater  than  the  fraction  — . 

The  multiplication  of  the  numerator  by  a  whole  number  may  also  be 

demonstrated  in  another  way.     —  is  taken  to  be  increased  c  fold,  and 

if  the  denominator  remains  unchanged,  while  the  numerator  is  multi- 
plied by  c,  the  number  of  parts  into  which  the  unit  is  divided  remains 
the  same,  but  the  number  of  those  parts  taken  is  increased  c  fold,  the 

oc 
new  fraction  —  must  then  be  c  fold  greater  than  the  old. 
b 

93.  II.  The  division  of  a  fraction  by  an  entire  quantity  is  effected 
by  dividing  the  numerator,  or  multiplying  the  denominator  by  the 
entire  quantity. 

For,  to  divide  --  by  c,  is  to  diminish  the  value  of  the  fraction  c  fold, 

and  if  the  result  of  the  division  of  a  by  c  is  q  ;  then  g'  is  c  times  smaller 

than  a,  and  y-  is  c  times  smaller  than  — .     Because,  while  the  parts  of 

the  unit  have  remained  unchanged,  c  fold  fewer  of  these  parts  have 
been  taken. 

The  division  can  also  be  effected  by  multiplying  the  denominator  by 
c,  for  then  the  parts  into  which  the  unit  is  divided  being  increased  c 
fold,  the  value  of  each  part  must  be  decreased  c  fold,  and,  of  course, 
when  the  same  number  of  these  c  fold  diminished  parts  are  taken,  as  at 
first,  the  result  must  be  c  times  smaller  than  before. 

94.  III.  The  value  of  a  fraction  is  not  altered  by  multiplying  the 

numerator  and  denominator  by  the  same  quantity.     The  fraction  --  is 

not  altered  in  value  by  multiplying  the  numerator  and  denominator  by 
c.     For,  to  multiply  the  numerator  by  c  is,  from  what  has  just  been 


ALGEBRAIC     FRACTIONS.  5H 

shown,  to  increase  the  value  of  the  fraction  c  fold,  and  to  multiply  the 
denominator  by  c  is  to  diminish  the  value  of  the  fraction  c  fold,  and, 
of  course,  the  fraction  has  undergone  no  change  of  value  since  it  has 
been  increased  and  decreased  equally. 

95.  IV.  The  viilue  of  a  fraction  is  not  changed  by  dividing  the 
numerator  and  denominator  by  the  same  quantity.  Because  the  two 
divisions  cancel  each  other,  and  leave  the  fraction  in  its  primitive 
condition. 

90.  Y.  If  the  same  quantity  be  added  to  the  numerator  and  denomi- 
nator of  a  proper  fraction,  the  value  of  the  fraction  will  be  increased. 

To  show  this,  it  will  be  necessary  to  show  that  if  the  same  quantity 
be  added  to  two  quantities  differing  in  magnitude,  the  smaller  of  these 
will  be  increased  more,  proportionally,  than  the  larger.  Take  the  num- 
bers 1  and  2,  add  1  to  both,  the  first  will  be  doubled,  but  the  second 
will  not  be ;  add  3  to  both,  the  first  will  be  increased  four  fold,  while 
the  second  Avill  only  be  2']  times  greater  than  before.  In  general,  let 
a  and  b  represent  the  two  quantities,  a  being  less  than  h-  add  a  to  both, 
the  first  will  be  doubled,  but  the  second  M'ill  not  be,  because  a  -\-  h  is 
loss  than  2/j.  Now,  when  the  same  quantity  is  added  to  the  numerator 
and  denominator  of  a  proper  fraction,  the  numerator  is  increased  more 
proportionally  than  the  denominator ;  the  number  of  parts  taken,  then, 
is  increased  without  their  being  an  equal  decrease  in  the  size  of  those 
parts.     Hence,  the  new  fraction  must  be  greater  than  the  old. 

97.  VI.  By  adding  large  quantities  to  the  numerator  and  denomi- 
nator, we  can  make  the  value  of  the  fraction  approximate  indefinitely 
near,  though  it  can  never  become  unity.  Thus,  add  1000  to  the  nume- 
rator and  denominator  of  i,  the  new  fraction  is  j^o^'  al"^os*^^>  though 
not  quite,  unity.  Now,  add  a  million  to  both  terms  of  the  fraction,  and 
the  result  will  be  indefinitely  near,  without  being  altogether  unity. 

98.  VII.  If  the  same  quantity  is  added  to  the  numerator  and  deno- 
minator of  an  im]m)per  fraction,  the  value  of  the  fraction  will  be 
decreased. 

For,  the  denominator  being  smaller  than  the  numerator,  will  be 
increased  more  proportionally,  and  the  value  of  the  fraction  must  be 
diminished. 

Take  |  as  an  example,  add  10,  and  the  fraction  becomes  |4  <C  f- 
It  is  plain  that  no  addition  to  the  numerator  and  denominator  can  ever 
reduce  the  fraction  as  low  as  unity. 
5* 


54  ALGEBRAIC    FRACTIONS. 


REDUCTION  OF  FKACTIONS. 

99.  Tlic  reduction  of  fractious  consists  in  simplifying  their  forms 
without  altering  their  value. 
Their  are  10  cases. 

CASE  I. 

To  reduce  a  simple  fraction  to  its  simplest  form. 

RULE. 

Strike  out  all  the  factors  common  to  the    numerator  and  denomi- 
nator^ and  the  result  will  he  the  fraction  reduced  to  its  lowest  terms. 

m                            1                    «'— ^' 
lake  as  an  example,  ^, 

We  know'  the  factors  of  a^  —  V^  to  be  (a  -\-  V)  (a  —  h),  hence, 


7/       („_5)^„+i) 


a  +  h  a  +  b 


—  h. 


The  expression  a,  +  b,  being  common  to  numerator  and  denominator, 
may  be  stricken  out,  since,  by  Article  95,  we  can  divide  both  terms  of 
the  fraction  by  a  +  ^  without  altering  its  value 

n  +  \ 

We  know,  by  Article  48,  that  n""  +  2n  +  1  =  (n  +  1)',  hence, 

n^  +  2n  +1         (n  +  1)  («  +  1)  ,    . 


«,  +  1 

(n 

+ 

1) 

■ 

^ 

Reduce 

x"  —  2ax  +  c 

i' 

X  —  a 

Ans.  X  —  a. 

4. 

Reduce 

Via'  — 121'- 

4rt  —   46  ■ 

Ans.  3  (a  +  h). 

5. 

Reduce 

Qa'^jcc  +  Q>ajcm 
36aa;  +  42aV 

Ans.   ac  +  m 

6   +7aV  ■ 

6.  Reduce 


7.  Reduce 


\.   Reduce 


ALaEBRAIC    FRACTIONS.  55 

4m^  —  \r? 


8 

(m  +  n) 

ac 

+  10  +  0" 

+  dc 

a 

+  h    +c 

+  d  • 

Zxy 

+  Q^y  - 

-  llxYt 

Ans. 


1  +  3x?/  —  ^xhjh, 
Ans. 


^bxy  +  oaxy  +  Vlx^f 

1  »;.<!-     ^  — . 

Zb  -\-  a     +  4xy 

CASE  II. 
100.  To  reduce  a  mixed  quantity  to  the  form  of  a  fraction. 

RULE. 

MuJtiph/  the  entire  qnantlti/  hy  the  denominator  of  the  fraction ; 
connect  the  product  xoith  the  numerator  hy  the  ajiprojyriate  sign  of  the 
numerator,  and  write  the  denominator  under  the  whole  result. 

Take  for  an  example,  a  +  ^,  the  result  is .      "\V'e   have   not 

altered  the  value  of  the  fraction,  since  the  division  hy  h  will  again  give 
us  a  +  — .     We  have,  in  fact,  only  multiplied  and  divided  a  by  the 


same  quantity. 

I. 

2.  Reduce 

,    x-a^ 

a  -\ 

a 

A               ^ 

Ans.    — . 
a 

3.  Reduce 

-X   i    -^-2-^  +  «^ 

.        x'  +  a' 

Ans. . 

a 

a 

4.  Reduce 

m^  —  n"- 
m  —  n  -\ . 

m  +  n 

A', 

ni  +  n 

5.  Reduce 

^^      ,    a^'  —  tax  +  x' 
X  —  a. 

Ans.  —  X. 

6.  Reduce 

2  ™2 

1+^+1-.- 

A       3-2x^ 

Ans. 

1  — x 

m 

ALGEBRAIC    FRACTIONS. 

7. 

Reduce 

X  +  h              .7c 

7  —  a  -\ .           Ans.   — 

c 

—  oc  +  X  -\-  b 
c 

8. 

Reduce 

l^  +  4c^  +  x' 

,      nx'  —  b' 

c'+  x'       ■ 

9. 

Reduce 

^         x'  —  2ax  +  3a' 
■ix  +  oa . 

X  +  a 

,        9ax  +  3.1 

Ans. 

X  +  a 

10.  Reduce  x  —  a  +  "^ .  Ans. 


11.  Reduce         b  +  2y  + 


X  -\-  a  X  -\-  a 

V  —  2hy—f 


h  +  y  '  Ans.  y. 


10    -P   1        11      ,   11         11    ,   11.'/  — 11^^^  +  11^-33;/+  22 

12.  Reduce  llx  +  11?/  —  11  -| — . 

x—y  +  1 

Ans.  11. 
CASE  III. 

101.  To  reduce  a  fraction  to  an  entire  or  mixed  quantity. 

RULE. 

Divide  the  mimerator  by  the  denominator  for  the  entire  part,  and 
place  the  remainder ,  if  any,  over  the  denominator  for  the  fractional 
part. 

x^  +  a^ 


Take 


X   -\-  a 


"We  have  a  right  to  divide  both  terms  of  the  fraction  (Article  95)  by 
the  denominator,  and  we  will  not  alter  the  value  of  the  fraction.  The 
reduction  in  Case  III,  is  nothing  more  than  dividing  both  terms  of  the 
fraction  by  the  denominator,  and,  since  the  new  denominator  is  unity, 
it  need  not  be  written. 

x^  -\-  a'^       x^  +  a?       x+  a 

Hence,  — ; =  — ; \ ; — , 

X  -{--a        X  +  a         X  +  a 

that  is,  both  terms  divided  by  x  +  a. 


ALGEBRAIC    FRACTIONS.  57 


mi  ,    •  —  ax  4-  (r  , 

The  result  is  x  + — - —  .      ,  ,    —  ax  +  a^ 

X  -\'  a  ,  or  simply,   x  -f 


8.  Reduce 

9.  Reduce 

10.  Reduce 


X  + 


«'  +  '^                              .                ^^ 
1.  Reduce  Ans.  a  -i . 


2.  Reduce  ~ ^" — '■ — —    '    ^.         Ans.  x  —  a  -\ 


a 

■^—2ax+  a' 

+  1 

X  —  a 

2  (ni^  —  n^) 

m  +  n 

l+x 

\—x 

\  —  x 

1   +  X 

1 

3.  Reduce  .  i       r.  ^  . 

Ans.  1  (in  —  n). 

4.  Reduce  ^    '   "■  Ans.  1  + . 

1  —  X 


2 
5.  Reduce  :; — : — .  Ans.  1 


\-\-x 


6.  Reduce  = — ; — .  Ans.  1  —  -— ^ — 
1  +  -^  1  +  a; 

1  X 

7.  Reduce  :j .  Ans.  1  + 


\  —  x  '    1—x 


7a'  +  l^ 
a'  +  6'' 

a  —b' 


Ans    7. 

Ans. 

7 

6i« 

«'  +  6«' 

A71S. 

a' 

+ 

ai  +  i'. 

CASE  IV. 
102.  To  develop  a  fraction  into  a  series. 

RULE. 

Divide  the  numerator  ly  the  denominator,  as  above,  and  continue 
the  division  as  far  as  may  be  required,  or  may  be  deemed  necessary. 


58  ALGEBRAIC    FRACTIONS. 

This  case  diffei-s  from  tlic  last  in  two  particulars,  1st,  It  includes 
only  fractious  which  cannot  be  reduced  to  entire  quantities.  2d.  The 
division,  instead  of  stopping  with  one  term  of  the  quotieut,Js  carried 
on  as  far  as  may  be  thought  proper. 


EXAMPLES. 


1.  Expand into  a  series. 


Ans.  1  +  2x  +  2x''  -f  2x^  +  &c.,  indefinitely. 


1 X 

2.  Expand into  a  series  of  five  terms. 

1  +  X 


Ans.  1  —  2x  +  2x2  —  2a;'  +  2x'  —  &c. 


3.  Expand  z into  a  series  of  six  terms. 


Ans.  1  —  X  -\-  x^  —  x^  +  X*  —  x^  +  &c 


4.  Expand  :; into  a  series. 

1  —  X 


Ans.   1  +  X  +  x''  +  x''  +  X*  +  &G. 


5.  Expand  —yj-ji  ^^^'^  ^  series. 


^       Qh'       Gh'       Gi" 
Ans.    I J   +  — r g   +  ^c. 


a  a  a 


6.  Expand =■  into  a  series. 

J_ns.  —  a;"'  —  x~^  —  cc"'  —  x-*  —  x-^  —  &c. 

7.  Expand r  into  a  series  of  five  terms. 

^         X  -\-  1 

Ans.  x-'  —  x-^  +  x-""  —  x-^  +  x-'  —  &c. 

x^ 

8.  Expand ^  .^         3  into  a  series. 

OC  uX     — j-  X 

Ans.  X  +  2a;2  +  ?^x^  +  4.x''  +  5.t"  +  &c. 


).  Expand 


X  ■\-y 

,  27/       %f   ,   2f       2f   ,    „ 


10.   Expand 


ALGEBRAIC    FRACTIONS. 
4 


Ans.  4x-'  —  4x-V  +  4x-'/  —  4.^"^^'  +  &t 


Remarks. 

103.  In  all  these  examples,  since  the  quotient  is  incomplete,  it  is 
plain  that  the  product  of  the  divisor  by  the  series  will  not  give  the 
dividend. 

104.  The  different  results  of  the  same  division  in  4  and  G,  3  and  7, 
show  that  the  series  may  be  changed  by  changing  the  order  of  the 
terms  in  the  divisor;  and  it  might  readily  be  shown  that  a  change  in 
the  order  of  the  terms  in  the  numerator  (when  there  is  more  than  one 
term)  will  produce  a  corresponding  change  in  the  series. 


CASE  V. 

105.  To  reduce  fractions  having  different  denominators  to  equivalent 
fractions  having  the  same  denominator. 


RULE. 

MuUipli/  each  numerator  into  all  the  denominators,  except  its  own, 
for  a  new  numerator,  and  all  the  denominators  together  for  a  common 
denominator  of  all  the  new  numerators;  or,  multipli/  each  numerator 
hy  the  quotient  arising  from  the  division  of  the  least  common  multij)le 
of  the  denominators  hy  each  denominator  respectively,  and  write  the 
products  thus  formed  for  new  numerators  over  the  least  common  multt- 
jile  as  a  common  denominator. 

Reduce  —  and  —  to  a  common  denominator. 
a  c 

By  the  first  rule  we  get  —  and  — ,  and  it  is  plain   that  these  frac- 
■^  "^     ac  ac  ^ 

tions  have  the  same  value  as  at  first,  since  each  has  been  multiplied  and 

divided  by  the  same  quantity. 

X  h 

Reduce  — ^  and  —  to  a  canimon  denominator. 


(SO  ALGEBRAIC    FRACTIONS. 

The  least  commoa  multiple  is  a^  •  the  quotient  of  this  common  multi- 
ple by  the  first  denominatoi'  is  1,  and  this  is,  then,  the  multiplier  of 

the  first  numerator.     The  multiplier  of  the  second  numerator  is  —  =  a. 

a 

Hence,  the  equivalent  fractions  are  —  and  -„ .      It  is   plain   that  each 
fraction  has  been  multiplied  and  divided  by  the  same  quantity. 

3.  Reduce  -^,  -V;  and  —   to   equivalent  fractions  with  a  common 
(r    cr  a 

denominator. 

X     ay    <j?h 


^"'-     .3'.,3^-3- 


4.  Reduce  J,  a,  -^,  and  —  to  equivalent  fractions  having  a  common 


denominator. 


25  Aah   hx^        ,  4?/ 

^«s-   7T)  -TT}  Try  and    .'^-. 

W   46  '  46  '  46 


5.  Reduce  h  — ,  r ^,)  and  ~r  to  equivalent  fractions  with 

m  +  ?i     {m  -j-  nj  4 

a  common  denominator. 

2  (m  -f  nf   4  (m  +  ',if  (to  —  n)   4  (y,^  +  a")        ,  (m  +  ?0'^ 
^"'-  4  (m  +  «)^'  4  (»7i  -^if         '  4  (m  +  nf         \{m  +  nf 

„    ^    ,         «    a;  +  a    «x  .  •     i     x  ^      x-         i      • 

6.  Reduce  — , ,  s  to  equivalent  tractions  havinc;  a  common 

X        a       a^ 

denominator. 

a?     ax^  +  a^x    ay? 

Ans.    -2-, ^ ,  -5-. 

o'-x  a^x  a^x 

7.  Reduce ,  — ,  ,'—  to  equivalent  fractions  havin<j  a 

in  —  n    h  X      '   c  ° 

common  denominator. 

hcx^  (m  —  n)  c'^x    (m  —  nybc     (m  —  n)I?x 

{in  —  n>)  hex'  (m  —  n)  hex'  (m  —  n)  hex'  (m  —  n)  hex' 

By  reducing  the  above  results  by  Case  I.,  we  will  get  back  the  ori- 
ginal fractions.  We  can  thus  verify  the  correctness  of  the  results 
obtained. 


ALGEBRAIC     FRACTIONS.  61 

CASE  VI. 
106.  To  add  fractional  quantities  top^ether. 

RULE. 

Reduce  the  fractions^  to  a  common  denominator,  if  necessary ,  and 
over  this  common  denominator  write  the  algebraic  sum  of  the  new 
numerators. 

Fractions,  which  have  different  denominators,  cannot  be  added  to- 
gether previous  to  reduction  to  the  same  denominator,  because  they 
represent  different  things.  Tims,  J  and  1  neither  make  i  nor  | ;  the 
first  indicates  that  one  of  the  three  parts  into  which  the  unit  has  been 
been  divided  has  been  taken ;  the  second  indicates  that  Ave  have  taken 
one  of  the  four  parts  into  which  unity  has  been  divided. 

Since,  therefore,  the  parts  taken  are  different  iu  magnitude,  they 
cannot  be  added  together.  It  would,  obviously,  be  just  as  proper  to 
add  a  peck  (the  fourth  of  a  bushel)  and  a  quart  (the  thirty-second  part 
of  a  bushel),  and  call  the  sum  two  pecks,  or  two  quarts,  as  to  add  two 
fractions  with  diflcreut  denominators. 


EXAMPLES. 
-f»ii=^  />  ,    —  C  .  ,  X  +   h  —  C 

1.  Add  —  H ,  and  together.  - 

a         a  a         ^ 

2.  Add  -,  -f  -  and  —  -. 

a         a  a 

3.  Add  —V-  to . 


4.  Add  X  —  a  -\ ^1 ^, 


a 

A  ns. 

X- 

+  ha  — 

ac 

a' 

Ans. 

4c 

:  +  ex  — 

2b 

2c 

3..;^  ■ 

—  2ax  + 

a"" 

r  —  2a  2  (x  —  a) 

_     .  ..2x       3x       4x       C)x  .       164a; 

6.  Mi'—l+J^to'-+J-'^. 
X  ^  +  y  c  n 

nc  (x'-'  — ;/)  -|-  bcxn  -f-  nx  {x  -f  y)^  —  mcx  (x  -\-  y) 

ncx  {x  4-  y) 
6 


02-  ALGEBRAIC     FRACTIONS. 

cc  —  y        m      ic  +  y         c 

^^^^    2mc  (x'  +  f)  +  (c"  +  m')  (a-''  —  .y') 
mc  (.x^  —  ?/^) 

8.  Add  4  +  4.x  to  J  +  ^.  ^n..  lL^+1). 

9.  Add  4  —  4a:  to  1  -  -|-.  ^«5.  mLlZ^). 

X  17  Tx —  1) 

10.  Add  4.r  —  4  to  ^  —  }.  Ans.  — ^ '-. 

4  4 

11.  Add  -^^ to  .T  —  a  H . 

6  X  -j-  a  X  —  a 

(x  +  a  +  bx  —  ha)  (a;^  —  a^)  +  2al^ 


Ans^ 


b  (x'  —  a^) 


Rcmarh. 


107.  When  none  of  the  terms  have  been  reduced  with  each  other, 
the  result  divided  by  the  common  denominator  ought  to  give  back  the 
original  series  of  fractions.  This  may  be  noticed  in  the  first  three 
examples. 

CASE  VII. 

108.  To  subtract  one  or  more  fractional  quantities  from  one  or  more 
fractional  quantities. 

RULE. 

Mahe  all  the  denominators  the  same,  if  the  fractions  have  not  a  com- 
mon denominator,  then  write  into  one  sum  the  numerators  of  the  minu- 
end, and  from  this  sum  take  the  algebraic  sum  of  the  numerator,  or 
numerators  of  the  subtrahend. 

Write  the  difference  over  the  common  denominator. 

The  same  reasons  which  show  that  we  cannot  add  fractions,  with 
different  denominators,  prove  that  we  cannot  subtract  fractions  with 
unlike  denominators.  The  fractions  represent  different  things  until 
reduced  to  a  common  denominator. 


ALGEBRAIC    FRACTIONS. 


1.  Subtract  -|-  from  y.  ^ns.  —  — . 

2.  Subtract  -J  from  — .  AnAns.  ^. 

6  2  t» 

3.  Required  the  diflference  between  x  and  -;^. 

X  X 

Ans.   -^  OT -^. 

4.  Take  the  difference  between  x  +  -—  and  x  —  -r-. 


Ans.   +  o,  or  —  a. 


^     „   ■,              a  4-  h         X    „         a  —  h 
5.  Subtract   — 1-  -;y  from  — - — 


i  ^>    ^^'   +   -^O  /7      ,         N 

Ans.  —  2  ;- — ,  or  —  (^  +  ^)- 


6.  Subtract  — f-  x  from  — - — 


Ans.  —  2  ^—.—  ,  or  —  (h  +  2x) 


7.  Subtract  2x  +  ^  -f  4  from  '^-. 


3(yH-6)  — 12a;  — 2a  — 24 
Ans.  --^^^^— ! — . 


8.  Subtract  ^^-^  from  2x4--^  +  "l- 
2  o 


Ans.   12-^  +  2«  +  24-8(y  +  &) 


^    A  ,  o -f  .r  -        a  —  X 

0.  Subtract  — - — from — ^ — . 


10.  Subtract  —^-  from  —f-. 

11.  Subtract  x  —  —  from  ' . 

6  7>l 


6 

Alls. 

(a  +  5.--) 
6 

a  -\-  5x 
Ans.    Ti — . 

,         ,  (x  +  c)  —  mhx  +  mx 

Ans.    b 1 • 

vib 


6Ei  A  L  Ci  E  B  15  A  1  C     1'  R  A  C  T  I  O  N  S  . 

CASE  VIII. 

109.  To  multiply  fractional  quantities  together. 

RULE. 

]f  the  quantities  are  mixed,  reduce  them  to  a  fractional  form ,  then 
mxdtiphj  their  numerators  together  for  the  numerator  of  the  product 
required,  and  their  denominators  together  for  the  denominator  of  this 
j^roduct. 

Let  it  be  required  to  multiply  -j-  by  — . 


To  multiply  —  by  c,  is  to  repeat  — ,  c  times ;  the  result  will  then 

plainly  be  — .  Because,  while  the  size  of  the  parts  into  which  the  unit 
has  been  divided  has  remained  the  same,  the  number  of  parts  taken  has 
been  increased  c  fold.     —  is  then,  obviously,  c  times  2;reater  than—-. 

But  we  were  not  required  to  multiply  —  by  c,  but  by  the  quotient  ari- 
sing from  the  division  of  c  by  d;  our  multiplier  has  then  been  d  times 
too  great,  and,  of  course,  the  product  is  d  times  too  great.  The  result, 
then,  must  be  corrected  by  dividing  by  d.     Hence,  the  true  product 

of  —  by  —  is  — „  in  accordance  with  the  rule. 
b         d       bd 


1.  Multiply  20^  +  1  by  I-.  Ans.  ^J^J^, 

.   2.  Multiply  ^^tf  by  ""-^ .  Ans.  ''—^. 


>.   3lultip]y ^—  by --.  Ans. 


a  ac 


4.  Multiply  -~^^  by ^.  Ans.  —. V- 

X -\- 1/  x—y  x^—y' 


ALGEBRAIC    FRACTIONS.  65 


5.  Multiply  ^44-— 'by^- f-*- 


A-hs    ^~^  "*"  ^~'^~^  ~  ^~^^~'  ~  ^~^ 


r^  4-  7/3 

6.  Multiply  '^   ^'^  by 


x'+U' 


{a 

2 

+  h)  {x- 

Ans.  1. 
Ans.    — . 

A 

7.  Multiply  7  +  ^^^  by  ^ ^, 

'  —  j/    '^  'y +i/■ 
^     ,,  ,  .  ,  ex  ,      1         // 

8.  Multiply  c  +  —  by  —  +  — . 

y     -^    c        r,/; 

9.  Multiply  L±i:!L-L/  byl^=:i:^-±^^.- 

c  (rt^  —  b^) 

_-,.,.-  a  +  6        ,         ac  —  he 

10.  Multiply  .,    ,  7  by.j ; . 

1— lGx^  +  %xy  —  f 

11.  Multiply  8  +  ^  —  c  by  ./:  —  ^  +  ^. 


Ans. 


4 
G4.r  +  4a.7-  —  8r.r  —  32a  —  2^^  +  5ao  +  l(3c  —  2c 


12.  Multiply   ^_ibyj+^, 


9*2  — 4/r        :>;^        V 
^"^-  -^1^*^^   4-9 


CASE  IX. 

110.  To  divide  fraetioual  quaiitities  by  each  other. 

RULE. 

If  tlie  quantity  to  Ic  divided  is  not  in  the  form  of  a  simple  fraction, 
reduce  it  to  that  form  ;  and.  if  the  quantity  to  he  used  as  a  divisor  is 
not  already  a  simple  fraction,  make  it  so.      Then  invert  the  terms  of 
the  divisor  and  proceed  as  in  the  last  case. 
G*  E 


m  ALGEBRAIC    FRACTIONS. 

Take,  as  an  example,  — -  to  be  divided  by  — . 

The  result  of  the  division  of  —  by  c  must  be  c  times  smaller  than—. 

It  will  then  plainly  be  — ,  because,  wliile  tlic  number  of  parts  taken  has 

remained  unaltered,  tlie  size  of  tlicsc  parts  has  been  diminished  c  fold, 
since  their  number  has  been  increased  c  fold.  But  we  were  not  re- 
quired to  divide  —  by  c,  but  by  the  quotient  of  c  by  d.     Our  divisor, 

then,  has  been  d  times  too  lar2;e,  and  the  result,  — ,  of  course,  d  times 

be 

too  small.     The  error,  then,  must  be  corrected  by  multiplying  by  d. 

Hence,  ^ — : — -  ■=  ^— ,  in  accordance  with  the  rule. 
h         d        be 

111.  The  demonstration  is  general,  and  is  applicable  when  the  divi- 
dend and  divisor  are  mixed  quantities,  or  made  up  of  several  fractions ; 
or,  when  either  dividend  or  divisor  is  a  mixed  quantity,  or  composed  of 
more  than  one  fraction.     For  the  fractions,  if  not  already  under  the 

form  of  -J-  and  -— ,  can  be  put  under  these  forms  without  difficulty. 
The  demonstration  in  the  last  case  is  general,  for  the  same  reasons. 


EXAMPLES. 

1. 

Divide  —  by  -^. 

Ans.  %x. 

2. 

Ax 
Divide  ~  by  5. 

Ans.  -^. 

3. 

Divide  -^  by  i. 

Ans.  4x. 

4. 

Divide  " — -j- —  by  - 

■  +  1 

8 

Ans.  2(a;  — 1). 

5. 

Divide  ^'' J  !:  + 

Hv^^ 

1 
1' 

^^  X  — 

Ans.  1. 

G. 

Divide  ^^^'^  by  - 
a    +6       ( 

y 

a'  —  b^' 

Ans.  (x—y)  (ci  —  h). 

ALGEBRAIC    FRACTION'S.  67 

r.  Divide  m  by  — ~  —  y.  Am.  -ir'^-^^7^—. 

II  a'  —  Ixy 

^.  Divide  !!^^  by  ^^-^V2x.  Am.        "^"^^ 


y  ■  n  {x^  +  y 

9.  Divide  1  +  ^  —  4:c  by  —  1  —  4-'  +  4.r. 

o  o  ^l«s.   — 1. 


10.  Divide  1  +  4^  _  4x  by  +  1  —  4  +  4.t. 

+  a;2  — 12x 


J^??s. 


3  —  a;^  +  Vlx 


^^     ^.  ..        2(x+y-)+a?j       -    ,       ,     Cr  — 7/)«  +  fl2^ 

11.  Divide      — ^^ — h  +  xhx  -T-. — -^ — — ; — . 

a  -^  4{x  +  y)  +  2ax 

Ans.   2^±^^^±^. 
cr\x  — y  +  ax 

10.  Divide      'i(^l+^'-l  +  .bj      U:r  +  y)  +  2a. 

X  —  y  +  ax 
Ans.   ^ . 

2a  +  X  +  2                 x        m  +  71       x 
lo.  Divide ^ 2o  +  -  by  -^, -. 

.        Ah{a  +  x  —  ch  +  1^ 

Ans.   — '-. 

m  -{-  n  —  2bx 

Rcmarlis. 

112.  The  quotient,  multiplied  by  the  divisor,  ought  to  give  the  divi- 
dend ;  and  in  the  last  case,  the  product,  divided  by  one  of  the  factors, 
ought  to  give  the  other  factor.  "We  are  thus  enabled  to  verify  our 
results. 

We  have  seen  that  both  the  numerator  and  denominator  of  a  fraction 
can  be  multiplied  by  the  same  quantity  without  altering  the  value  of 
the  fraction.     Hence,  by  multiplying  the  term  of  a  fraction  by  minus 

X 1 

iinity,  the  signs  of  these  terms  may  be  altered.     Thus,  " — - —   may  be 

•xi      1  —  x     -\-  a  —  a    .,         _  —  a        a 

written  — ; may  be  written  =-,  cvc.     Hence,  j  =  — . 

—  0      —  0  +6  —  bo 

That  is,  the  quotient  of  two  negative  quantities  is  positive. 
So,  - — T,  or, J  = -,  read  minus  the  fraction  — . 

+    O  —  Ob  0 


ALGEBRAIC    FRACTIONS. 


CASE  X. 


113.  To  reduce  a  compound  fraction  to  its  lowest  terms. 

Wc  have  seen  tliat  a  simple  fraction  can  often  be  reduced  by  remov- 
ing the  factors  common  to  the  numerator  and  denominator.  When  the 
common  factors  of  a  compound  fraction  can  be  detected  by  inspection, 
we  have  but  to  remove  them,  and  the  fraction  is  reduced  to  its  lowest 
terms.     Thus, 

a'^  +  ah  —  ac 


um  -\-  an 
can  be  reduced  to  its  lowest  terms  by  the   removal  of  the   common 
factor  a. 

But  the  common  factor,  or  common  factors,  of  a  compound  fraction 
cannot  always  be  detected  by  inspection,  and  some  process  becomes 
necessary  to  discover  them.  This  process  is  called  finding  the  (jrealest 
common  divisor. 

114.  The  greatest  quantity  that  will  divide  two  or  more  quantities, 
is  their  greatest  common  divisor. 

When  the  greatest  common  divisor  of  the  numerator  and  denomina- 
tor of  a  compound  fractions  is  found,  it  can  be  reduced  to  its  lowest 
terms  by  dividing  by  this  divisor. 

115.  The  greatest  common  divisor,  though  most  usually  obtained  in 
order  to  reduce  compound  fractions,  is  also  frequently  found  between 
(quantities  not  written  in  the  fractional  form. 

IIG.  We  will  then  explain  the  method  of  finding  the  greatest  com- 
mon divisor  between  two  polynomials,  without  regarding  them  as  nume- 
rator and  denominator  of  a  compound  fraction. 

117.  The  determination  of  the  greatest  common  divisor  of  two  poly- 
nomials depends  upon  two  principles.  1st.  The  common  divisor  of 
two  polynomials  contains,  as  factors,  all  the  common  factors  of  the 
two  polynomials,  and  does  not  contain  any  other  factors.  For,  let  A 
and  B  be  the  two  polynomials,  and  D  their  greatest  common  divisor. 
Now,  it  is  plain  that  any  quantity  C,  made  up  of  a  part  of  the  common 
factors  of  A  and  B,  would  divide  both  quantities,  and  would,  therefore, 
be  a  common  divisor ;  but  C,  multiplied  by  the  remaining  factors  com- 
mon to  A  and  B,  would  still  divide  them.  Hence,  C  would  be  a  divi- 
sor, but  not  the  greatest  divisor  of  A  and  B. 

Again,  any  quantity  M,  made  up  of  all,  or  a  part,  of  the  common  fac- 


ALGEBRAIC     FRACTIONS.  69 

tors  of  A  and  B,  and  containing  also  a  factor,  d,  not  common  to  the 
two  polynomials,  will  not  divide  them.  For,  if  we  proceed  to  the  divi- 
sion, the  common  factors  of  A,  B,  and  M  will  strike  out,  and  leave  the 
factor  (1  as  a  denominator  of  the  reduced  polynomials.  Thus,  ad  will 
not  divide  a^  -f  ah  and  ac  +  am  ;  for,  though  it  contains  their  common 
factor,  a,  it  also  contains  a  factor  d,  not  common.     The  result  of  the 

division  will  be       ,      and  - — - — .     The  greatest   common  divisor,  D. 
a  d 

then,  contains  all  the  common  factors  of  A  and  B,  and  contains  nu 

other  factors. 

118.  2d.  The  greatest  common  divisor  of  two  polynomials,  A  and  B, 
will  enter  into  the  successive  remainders  which  arise  from  dividing  A 
by  B,  and  B  by  the  remainder,  and  the  second  remainder  by  the  first, 
aud  so  on,  until  there  is  no  remainder. 

For,  denote  by  A'  and  B'  the  quotients  arising  from  dividing  A  and 
B  by  the  greatest  common  divisor,  D ;  then  A  =  A'D,  and  B  =  B'D. 
Divide  A  by  B ;  or,  what  is  the  same  thing,  divide  A'D  by  B'D,  aud 
call  the  quotient  Q. 

Thus,  A'D      I  B'D 

QB'D      Q 

(A'  — QB')  D  =  1st  Bemaindcr  =  MD. 
by  making  A'  —  QB'  =  IM. 

We  see  that  the  first  remainder  contains  the  greatest  common  divisor; 
and  as  it  is  also  in  the  divisor,  wc  have  a  right  to  seek  it  between  these 
polynomials. 

Now,  divide  B'D  by  the  remainder,  and  call  the  new  quotient  (7, 

and  we  get 

B'D       I  MD 

Q':md     q' 

(B'  — Q'M)  D  =  2d  Kemaindcr  =  XD, 
representing  B'  —  Q'M  by  it. 

Wc  see  that  the  greatest  common  divisor  enters  also  into  the  second 
remainder;  and  as  it  is  also  in  the  first  remainder,  we  have  a  right 
to  seek  it  between  these  two  remainders.  Divide  MD  byND;  the 
remainder,  if  any,  will  still  contain  the  greatest  common  divisor. 

119.  Let  us  apply  these  principles  in  finding  the  greatest  common 
divisor  of  the  polynomials,  A  and  B. 

First  arrange  them  with  respect  to  a  certain  letter,  and  regard  that 
polynomial  which  contains  the  highest  power  of  this  letter  as  the  divi- 


70  ALGEBRAIC     FRACTIONS. 

dend,  and  the  other  polynomial  as  the  divisor.  Next,  examine  if  A 
(which  we  suppose  the  dividend)  contains  a  factor  common  to  all  its 
terms,  but  not  common  to  all  the  terms  of  B.  By  the  first  principle, 
this  factor  can  constitute  no  part  of  the  common  divisor,  and  may  be 
suppressed.  It  is  not  absolutely  necessary  to  suppress  it,  because  it 
will  disappear  in  division,  and  not  appear  in  the  remainder,  and,  there- 
fore, by  the  second  principle,  cannot  make  part  of  the  greatest  common 
divisor.  But  if  there  is  a  factor  common  to  all  the  terms  of  B,  and  not 
common  to  those  of  A,  it  must  be  suppressed.  For  A  would  not  be, 
divisible  by  B  until  it  had  been  multiplied  by  the  common  factor  of  B, 
and  the  multiplication  would  make  this  factor  common  to  A  and  B ; 
and,  hence,  by  the  first  principle,  it  must  be  common  to  the  divisor. 
The  greatest  common  divisor  would  then  contain  a  factor  which  did  not 
originally  belong  to  A. 

120.  If  there  is  a  factor  common  to  A  and  B  which  can  be  detected 
by  inspection,  it  may  be  divided  out  and  set  aside,  as  making  part  of 
the  greatest  common  divisor. 

121.  The  next  step,  after  setting  aside,  or  suppressing  factors  seen 
by  inspection,  is  to  divide  A  by  B.  If  the  coefiicient  of  the  first  term 
of  A  is  not  divisible  by  the  coefiicient  of  the  first  term  of  B,  it  may  be 
made  divisible  by  multiplying  all  the  terms  of  A  by  the  coefiicient  of 
the  first  term  of  B,  or  by  any  other  quantity  that  will  make  the  division 
possible.  This  multiplication  will  not  efi"ect  the  result,  because  the 
factor  introduced  into  it  will  disappear  in  the  division. 

122.  The  remainder,  after  division  of  A  by  B,  will  contain  the  great- 
est common  divisor ;  and,  as  B  also  contains  it  by  hypothesis,  we  have 
a  right  to  seek  it  between  B  and  the  remainder.  The  second  remain- 
der, if  any,  contains  D  also,  but  its  coefficient  or  multiplier  is  smaller 
than  in  the  first  remainder;  and  so  tlie  coefiicient  of  J),  in  each  re- 
mainder, is  smaller  than  in  the  preceding,  until  it  finally  becomes  unity. 
"When  this  final  remainder  is  used  as  a  divisor,  it  will  go  as  many  times 
in  the  preceding  divisor  as  there  are  units  in  the  coefficient  of  D  in 
that  divisor,  and  there  will  be  no  remainder. 

123.  When,  therefore,  wc  get  a  remainder  zero,  we  conclude  that 
the  last  divisor  is  D  X  1,  or  D  itself. 

124.  If  the  given  polynomials  have  no  common  divisor,  we  can  dis- 
cover the  fact  by  the  same  tests  that  show  when  one  polynomial  is  not 
divisible  by  another.  We  will  find  either  that  the  letter,  according  to 
which  the  arrangement  has  been  made,  has  disappeared  from  some  of 


ALGEBRAIC    FRACTIONS.  71 

the  remainders,  or  it  enters  to  a  higher  power  in  that  remainder  which 
is  used  as  a  divisor  than  in  the  one  used  as  a  dividend. 

The  preceding  principles  and  demonstrations  for  finding  the  greatest 
common  divisor  lead  to  the  following 


RULE. 

125.  Arraivje  the  two  polynomials  with  reference  to  a  certain  letter, 
as  in  division;  use  that  one  tchich  contains  the  highest  j^oicer  of  this 
letter  as  the  dividend. 

2d.  Next,  set  aside  the  factors,  if  any,  common  to  dividend  and 
divisor,  as  part  of  the  greatest  common  divisor,  and  suj^ress  those  fac- 
tors tohich  are  common  to  the  one  and  not  to  the  other. 

3c7.  Prepare  the  dividend,  if  necessary,  for  division,  hy  multiplying 
hy  any  quantity  that  loill  make  its  first  term  divisible  hy  the  first  term 
of  the  divisor ;  divide  the  one  polynomial  hy  the  other,  and  suppress 
in  the  remainder  any  factor  that  may  he  common  to  cdl  its  terms  and 
not  common  to  those^of  the  divisor. 

Aih.  Use  the  remainder  so  reduced  as  a  divisor,  and  the  last  divisor 
as  a  dividend,  and  proceed  c(s  before;  and  continue  in  tliis  manner, 
using  eadh  successive  remainder  as  a  divisor,  and  the  last  divisor  as  a 
dividend,  until  tliere  is  no  remainder,  or  until  it  is  evident  that  the 
polynomials  have  no  common  divisor. 

„    ,         a'^cb  —  a^c)/  +  a^b.r  —  a^i/.r       . 

lieduce  — '^ '-^—  to  its  lowest  terms. 

mcb  +  mcb  +  mbx  +  myx 

a^  is  common  to  the  numerator  and  not  to  the  denominator,  m  is  com- 
mon to  the  denominator  and  not  to  the  numerator.  These  factors  must 
then  be  suppressed,  as  constituting  no  part  of  the  greatest  common 
divisor. 

Then,  ch  —  cy  +  hx  —  yx  \  cb  -f  cy  +  hx  -\-  yx 

cb  -\-  cy  -f  hx  +  yx  1  Quotient. 

—  'Icy  —  "lyx  =  —2y{c-\-  x). 

Suppress  —  2y,  a  factor  common  to  the  remainder  and  not  to  the 
divisor. 

Tlien,  cb  +  cy  -\-  hx  -\-  yx  \  c  -\-  x 

cb  +  bx ^~-\-  y    2d  Quotient. 

cy  +  yx 
cy  +  yx 
0  +  0. 


72  ALGEBRAIC    FRACTIONS. 

Hence,  c  -}-  x  is  tlic  greatest  common  divisor,  and  the  reduced  frac- 
tion obtained  by  dividing  both  terms  of  the  fraction  by  the  G,  C,  D,  is 

«'  (&—;/) 
m  (6  +  v/)' 

The  factor  suppressed  in  the  first  remainder  might  have  been  +  2y, 

instead  of  —  2?/.     The  greatest  common  divisor  then  would  have  been 

—  c  —  X.     And,  in  general,  the  two  polynomials  have  two  greatest 

common  divisors,  differing  in  the  signs  of  all  their  terms.     The  reason 

„  ,  .    .     ,    .  .        Ad        —  Ad     .  ,.  .  , 

01  this  IS  obvious,  since  ^f—  =  ^  ,  the  common  divisor  to  the  two 

JJd         —  lj(t 

terms  of  the  fraction,  may  be  either  +  (7,  or  —  d. 

2.  Find  the  greatest  common  divisor  of  x^  +  4x^  +  bx  +  2,  and 
2x^-{-  Sx  -f  1.  Prepare  for  division  by  multiplying  the  first  polyno- 
mial by  4,  the  square  of  the  coefficient  of  the  first  term  of  the  second 
polynomial. 

Then,  Ax''  +  ICr^  +  20:c  -f  8  I  2:c"  -f  3x  +  1 

4.x=^  +     Cx^  +  2x  I        2x  4^5  Quotient. 

10^2  +  18a;  +  8 
lO.x^'  +  15x  +  5 


3:c  +  3  =  3  (x  +  1).     Remainder. 

Suppress  the  common  factor,  3,  of  the  remainder,  and  continue  the 
division. 

2^2  +  Sx4-1  \    x  +  1 

2x^  +  2x  2x  +  1.     2d  Quotient,     a'  -f  1  =  G.C.D. 

X  +  1 
X  +  1 

126.  If  we  had  multiplied,  to  prepare  for  division,  by  the  first  in- 
stead of  the  second  power  of  the  cocfiicient  of  the  first  terra,  we  would 
only  have  found  one  term  in  the  quotient  by  the  first  division,  and  it 
would  have  been  necessary  to  multiply  the  remainder  by  the  coefficient 
of  the  first  term  to  obtain  a  second  term  of  the  quotient.  It  is  easier 
to  prepare  the  polynomial  by  multiplying  by  the  second  power  of  the 
coefficient  of  the  first  term. 

3.  Reduce ,--^ ; = to  its  lowest  terms. 

6x^  —  4x-  +  1 

Multiply  the  numerator  by  3'  =  27,  to  prepare  for  division. 


ALGEBRAIC    FRACTIONS.  73 

27x^  +  WSz'  —  Slx'  —  'ZlOx  +  210  |  Hx^  —  ix  +  1 


21x'—   36x«+    9x' 

9x^ 

+  48x  + 

34 

144a;'- 
144a;'- 

—  90x' 

—  192.r^ 

—  270x  +  216 

'+    48a; 

102x2 
102a;^ 

— 318.r  +  216 
—  136.C  +    34 

Suppress  — 

182. 

—  182a; +  182: 

=  - 

-182  (a- - 

■!)• 

Then, 

3a-2  —  4x  +  1  1 

a;  • 

-1 

3a;^  — 3x 

Sx- 

-1 

—  a;  +  l 

—  a;  +  l 

0  .  0 


+  5x'  -^ 


ox  —  1 

127.  In  the  above  examples,  the  difference  of  the  exponents  of  the 
arranged  letter  being  two,  we  multiply  by  the  square  of  the  coefficient 
of  the  first  term  of  the  divisor  to  prepare  the  dividend  for  division.  It 
could  have  been  prepared  by  multiplying  by  3,  but  we  would  only  have 
gotten  one  term  in  the  quotient,  and  there  Avould  have  been  two  re- 
mainders to  prepare  for  division. 

In  general,  when  preparation  for  division  is  necessary,  we  save  time 
by  multiplying  the  dividend  by  the  coefficient  of  the  first  term  of  the 
divisor,  raised  to  a  power  one  greater  than  the  difference  of  exponents 
of  the  arranged  letter  in  the  first  terms  of  the  two  polynomials. 

4.  Find  the  greatest  common  divisor  of 

«''  —  a^x  +  ax^  —  x^,  and  a'  —  rrx  +  ax-  —  x^. 

Ans.   a  —  X. 

_     _    -        cx^  -f  ex*  +  or  —  x^  —  x^  —  xi/^ 

5.  Reduce  -^ ,       '^ , ^—, — . ^,.  ■ 

fa-'  —  ex    +  CI/  —  X    -\-  x  —  Xi/ 

Ans.  GCD  c  —  x. 

And  reduced  fraction,  -= s —„. 

sr — x^  +  1/^ 

7 


74  ALGEBRAIC    FRACTIONS. 

6.  llcduce  '-^^,      ^'^„      '^"^,        !^,  to  its  lowest  terms. 


Ans.  GCDx^  — ?/l 


Reduced  fraction,         ^"^  +  ^^  +  f '  +  < 

c  +  2 


-    -r,    ■,        '"'f'  —  2ama;  +  mx^  +  a^  —  2ax  +  x^      .     . 

7.  Ecduco 5 — -r 5 to  its  lowest  terms. 

na^  —  Zanx  +  nx^ 


Reduced  fraction, 


Ans.  GCB  a^  —  2ax  + 
m  +  1 


8.  Find  the  GCD  of  3.^*=  —  3/  and  40;=*  +  4/. 

Ans.  x^  +  y^. 

9.  Find  tlie  GCD  of  x"  ■\-  x"  —  ax''  +  a-x  —  «j;  +  a^  and  x''  —  x^ 
—  ax}  +  a^x;  +  ax  —  cr. 

J-jis.  £c^  —  ax  -\-  a-. 

10.  Find  the  GCD  of  a:'«  —  ^'^  and  x^  +  x^  +  xif  +  y\ 

Ans.  .c^  +  ,y^. 

11.  licduce  ^r — - — -— — to  its  lowest  terms. 

X?  +  \ax^  +  Soil-  +  2a^ 


Reduced  fraction, 


Ans.  GCD  3?  +  2ax  +  a\ 
x^  +  2o.r  +  a" 
X  +  2a 


12.  Reduce  -5 r ; , ,  to  its  lowest  terms. 

x-  —  X  +  x//~^  —  ,y     +  •'*■     —  -^ 

Ans.  GCD  .r-2  +  3/-'  4-  X. 

Reduced  fraction, =. 

a;  —  1 

13.  Find  the  GCD  of  .7;^  +  x^_//  +  1+^  +  1/  +  yx-^  and  x^  +  yx^ 

+  ^'^'  +  y  +  J/=  +  ^• 

d[?!S.   a;''  +  ^. 

^  ,    ^    .  :^V  +  2*"  +  3a;  +  o-y  +  1     ^    .     . 

14.  Reduce  - — -^ = -. — \  ,  ,^    to  its  lowest  tenus. 

a;^+ 2x  +  1 +yx-' +  ?/' +  2y 

^Ihs.  GCD  a;-'  +  y  +  2. 
Reduced  fraction,  ^— — ^. 


ALGEBRAIC    FRACTIONS.  75 

Eemainder  in  this  example,  2x  —  y  +  1  —  ary  —  jfx  =  2a;  +  1 
->r  xy  —  2xy  —  y  —  y'x  =  X  (x"'  +  y  -\'T)—yx  (a;-'  +  2  +  y)  = 
(x—yx)(x-'  +  2+y-). 

Suppress  factor,  cc  — yx. 

15.  Reduce  :; ^— — ; ~^—, ^  to  its  lowest  terms. 


1  +  jy-'  + 


Reduced  fraction,  ' 

a;-'  +  y- 


Ans.  GCD  x  +  7f. 


Eemarhs. 

128.  The  last  two  examples  show  the  importance  of  suppressing  tho 
factor  common  to  the  remainder  and  not  common  to  the  divisor. 

Greatest  common  divisor  of  three  or  more  polynomials. 

129.  The  foregoing  principles  can  be  readily  extended  to  finding  the 
greatest  common  divisor  of  three  or  more  polynomials.  It  is  evident, 
that  if  we  use  one  of  the  polynomials  as  a  divisor,  and  divide  all  the 
others  by  it,  that  the  remainders,  if  any,  must  contain  the  common 
divisor  of  all  the  polynomials.  Suppress,  in  each  of  these  remainders, 
the  factors  not  common  to  the  polynomial  used  as  a  divisor,  and  take 
that  remainder  which  contains  the  lowest  power  of  the  arranged  letter 
as  the  new  divisor,  and  divide  the  other  remainders  by  it;  and  con- 
tinue in  this  way  until  we  get  an  exact  divisor;  it  will  be  the  GCD 
sought.     To  illustrate  by  an  example,  find  the  GCD  of 

a-2  +  ax  —  2.V  —  2r(,  :r^  —  .>•  —  2,  and  ./'^  —  2./-'  —  4.r  +  8. 

Use  x^  —  X  —  2  as  the  divisor,  the  remainder  will  be 

ax  —  X—  2a  +  2,  or  a  (x  —  2)  —  {x  —  2)  =  (./  —  !)  (x  —  2),  0 
and  —  a;-  —  2a;  +  8 

Suppress  the  factor,  a  —  1,  in  the  first  remainder,  and  the  result, 
X  —  2,  will  exactly  divide  the  other  two  remainders ;  hence,  x  —  2  is 
the  greatest  common  divisor  of  the  given  polynomials. 

Again,  take 
x''+ax  +  x  +  a,  x''  +  Sx^  +  ?jx^l,  and  x^  +  2x''  —  ax^  —  2ax  -f  .x  — a. 


76  ALGEBRAIC    FRACTIONS. 

Use  the  first  as  the  divisor;  tlie  three  remainders  will  he 

0,  —  2«  (,r  +  1)  -f  :f  +  1  +  d'  {x  +  1),  aud  1a^  {x  +  1)  —  2a{x-\-  1), 

or  0,  and  (..•  +  1)  (1  +  a^  —  2a),  and  (.«  +  1)  C-«'  —  2fl). 

Hence,  x -\- I  =  GCD. 

3d.  Find  the  GCD  of  a''  —  x\  a?  +  2ax  +  a'  aud  (c"  +  3«lr  + 
3ax2  4-  .-r^ 

J.HS.  a  +  X. 

4th.  Find  the  GCD  oix^  —  -f,  x^  —  f,  x''  —  i/'%  and  .r^'  —,?/=•. 

^Ins.  .x^  —  1/^. 

5th.   Find  the  GCD  of  .r-  —  1,  :r.^  —  l,x''  —  2x  +  1,  and  x''  —  Sx' 

+  nx  —  i. 

Alls.  X  —  1. 

130.  It  will  be  seen  that  the  GCD  of  any  number  of  polynomials,  as 
well  as  for  two,  may  have  its  sign  changed.  Thus,  the  last  GCD  may 
he  X  —  1,  or  1  —  X,  and  so  for  the  others. 

LEAST   COMxMON   MULTIPLE. 

131.  The  least  common  multiple  of  two  or  more  quantities,  is  the 
least  quantity  that  they  will  exactly  divide.  Thus,  the  least  common 
multiple  of  2,  4,  and  6,  is  12;  of  a^,  a,  and  h,  is  a-h;  of  a^  +  ah,  I  and 
^/,  is  a^h^  +  «Z/,  &c.  It  is  plain  that  the  product  of  all  the  quantities 
will  be  a  common  multiple  of  these  quantities;  that  is,  it  will  be  exactly 
divisible  by  them.  Thus,  2x4x6  =  24  is  a  common  multiple  of  2, 
4,  and  6,  but  it  is  not  their  least  common  multiple.  In  like  manner, 
a'^h  is  a  common  multiple  of  a^,  a,  and  h,  but  not  their  least  common 
multiple.  Multiply  a  common  multiple  by  anything  whatever,  the  pro- 
duct will  still  be  a  common  multiple.  Hence,  there  may  bo  an  infinite 
number  of  common  multiples,  but  there  can  be  but  one  least  common 
multiple. 

132.  No  quantity  will  be  divisible  by  another,  unless  it  contains  all 
the  factors  of  the  second  quantity.  So,  no  quantity  will  be  divisible 
by  two  or  more  quantities,  unless  it  is  divisible  by  each,  and  by  all  the 
factors  of  these  quantities.  To  be  the  least  common  multiple  of  the 
given  quantities,  it  must  contain  no  more  factors  than  they  contain;  and 
these  factors  must  not  enter  to  higher  powers  than  in  the  given  quan- 
tities.    Thus,  abc  is  not  the  least  common  multiple  of  a  and  h,  because 


ALGEBRAIC    FRACTIONS.  77 

it  contains  a  factor,  c,  that  they  do  not  contain.  Neither  is  a^b  the 
least  common  muhiple^  because  the  factor  a  enters  to  a  higher  power 
than  in  the  given  quantities.  The  expressions  abc,  and  a^b,  are  multi- 
ples, but  not  hast  common  multiples. 

133.  If  the  given  quantities  contain  a  common  factor,  this  must 
enter  into  the  least  common  multiple  raised  to  the  highest  power  ti) 
which  it  is  raised  in  the  expressions  to  be  divided,  but  it  must  not  be 
repeated. 

134.  It  must  enter  to  the  highest  power,  else  the  multiple  which  we 
formed  would  not  be  divisible  by  the  expression  containing  the  highest 
power  of  the  common  factor,  and  it  must  must  not  be  repeated,  else  the 
multiple  would  not  be  the  least  common  multiple. 

135.  From  these  principles  we  derive  the  following 

RULE. 

Decompose  the  given  quantities  into  their  prime  factors,  form  a  pro- 
duct composed  of  all  the  factors  not  common,  and  of  the  highest  powers 
of  the  common  factors,  talcinfj  care  to  let  no  factor  enter  more  than 
once.  The  product  so  formed  will  le  the  least  common  multiple 
required. 

Form  the  least  common  multiple  of  ax^,  a^x,  bx*,  and  a'. 

Decomposing,  we  have  a.x^,  a^.x,  h.x*,  and  a'. 

Hence,  least  common  multiple,  a'^.x*.b  =  aVjx*. 

It  is  plain  that  the  common  factor,  a,  must  enter  to  the  highest 
power  (the  third)  to  which  it  enters  in  one  of  the  given  quantities, 
else  the  multiple  would  not  be  divisible  by  that  quantity.  It  is  also 
plain  that  x  must  enter  to  the  highest  power,  and  that  the  factor  b, 
not  common,  must  also  form  part  of  the  least  common  multiple,  other- 
wise, the  expression  coutaiuiug  b  would  not  be  a  divisor. 

Form  the  least  common  multiple  of  2,  8,  3,  9,  a,  a^,  and  5b. 

Decomposing  into  factors  we  have  2,  2',  3,  3^,  a,  o?,  h.b. 

Hence,  2'.3^.alZ».5  =  360a^i  is  the  least  common  multiple  required. 

By  inspecting  the  result,  2^,  3^,  ha^.b,  it  is  plain  that  it  is  the  least 
common  multiple.     It  is  divisible  by  2,  because  it  contains  a  factor  2^ ; 
7* 


78  ALGEBRAIC    FRACTIONS. 

by  8,  because  it  contains  2^;  by  3,  because  it  contains  3^,  &c.  More- 
over, it  is  the  least  product  that  will  divide  the  given  quantities,  for  it 
is  made  up  of  the  least  factors  that  will  fulfil  the  required  conditions. 
The  number  2  need  not  be  raised  to  the  third  power  to  divide  2,  but 
it  must  be  to  divide  8 ;  so  3  need  not  be  raised  to  the  second  power  to 
divide  3,  but  it  must  be  to  divide  9. 

We  see,  too,  that  no  factor  has  been  taken  more  than  once. 

3d.  Find  the  least  common  multiple  of  (a^  +  ax^^,  9a'',  21,  and  7ah. 
A}is.  3^7  («  +  .t')  a^h  =  63  (a  +  x")  a^h. 

4th.  Find  the  least  common  multiple  of  ,t^  —  ?/,  x -\- y,  7x  —  7y, 
and  x'^  —  7/\r. 

Ans.  1x  (.c^  —  ?/^). 

5th.  Find  the  least  common  multiple  of  ?Hi^h\  4  Qx  +  If,  IQcc^lr, 
and  40  {x  +  l)/>. 

Ans.  3.2^5  (.r  +  Yfa^J?  =  240  (x  +  1)V/A 

6th.  Find  the  least  common  multiple  of  2a,  oa^,  4i,  5//,  Gc,  7c^,  8d\ 
dJ,  and  lOahcd. 

Ans.  2\B'.b.7a%'c\I'  =  2520a'b\M\ 


CoraUarj/. 

136.  The  least  common  multiple  of  two  or  more  quantities  can  also 
be  found  by  dividing  their  product  by  the  greatest  common  divisor. 
For,  in  the  division  of  the  product  of  the  quantities  by  their  greatest 
common  divisor,  the  lowest  powers  of  their  common  factors  alone  are 
divided  out,  and  the  quotient  is  made  up  of  the  factors  not  common, 
multiplied  by  those  that  are  common,  raised  to  the  highest  powers  to 
which  they  enter  in  any  of  the  given  quantities,  which,  as  we  have 
seen,  is  the  composition  of  the  least  common  multiple. 

137.  Conversely,  if  we  know  the  least  common  multiple  of  two  or 
more  quantities,  we  can  find  their  greatest  common  divisor  by  multi- 
plying the  quantities  together,  and  dividing  their  product  by  the  least 
common  multiple.  For,  let  A,  B,  and  C  denote  the  quantities,  D,  their 
greatest  common  divisor,  and  L  their  least  common  multiple.     Then, 

since  =  L,  -^ —  =  D.     This  relation  between  the  greatest  com- 

mon divisor  and  the  least  common  multiple,  enables  us  to  verify  our 
results  iu  finding  either. 


ALGEBRAIC    FRACTIONS. 


LEAST  COMMON  MULTIPLE  OF  FRACTIONS. 

138.  It  is  often  important  to  find  the  least  common  multiple  of  frac- 
tions, and  as  the  rule  for  finding  it  for  entire  quantities  fails  in  this 
case,  it  becomes  necessary  to  demonstrate  another  nxle. 

Let  us  take  the  fractions 

"We  arc  required  to  find  the  least  quantity  that  they  will  exactly  divide, 
and  give  entire  quotients.  Now,  to  divide  by  a  fraction,  is  to  divide 
by  the  numerator,  and  multiply  by  the  denominator.     The  quantity, 

then,  that  will  be  divisible  by  one  of  the  fractions,  as  -— ,  must  be  divi- 
de 

sible  by  a^,  or  it  will  not  be  divisible  after  it  has  been  multiplied  by  hc^. 
And,  as  the  same  reasoning  may  be  extended  to  the  other  fractions,  the 
quantity  sought  must  evidently  be  divisible  by  each  of  the  numerators ; 
and,  in  order  to  be  the  least  quantity  that  will  fulfil  this  condition,  it 
must  be  the  least  common  multiple  of  the  numerators.  But,  since  the 
denominators  are  to  be  multiplied  by  this  least  common  multiple  of  the 
numerators,  it  is  plain  that  a^h  cannot  be  the  quantity  sought.  For, 
rt^i,  the  least  common  multiple  of  the  numerators,  divided  by  any  quan- 
tity that  will  exactly  divide  the  denominators,  will  be  a  smaller  quantity 

than  aH>  itself.  For  instance,  —  will  be  divisible  by  the  given  quan- 
tities, and  be  a  smaller  quantity  than  a^b.  It  is  plain,  too,  that  a^b, 
divided  by  the  greatest  quantity  that  will  divide  the  denominators,  will 

still  be  smaller  than  -— ,  and  will  be  the  smallest  quantity  that  will  be 

exactly  divisible  by  the  given  quantities.  Hence,  -j.  is  the  least  com- 
mon multiple  of  the  fractions 

6?'     P?^"'^^- 


RULE. 

139.  Find  the  least  common  multiple  of  tlie  mimerators,  and  divide 
it  by  the  greatest  common  divisor  of  the  denominators ;  the  fraction  so 
formed  tcill  be  the  least  common  multiple  of  the  (jiven  fractions. 


80  A  li  G  E  B  R  A  1  C     F  R  A  C  T  I  0  N  S  . 

2.  Find  the  least  comiuou  multiple  of  -f^,  4^,  and  K-. 

3.  Find  tlic  least  common  multiple  of 


Arts.   V- 


8  (a'  +  ax') 


-^3—,—-^—,  and-. 

Ans. 

4.  Find  the  least  common  multiple  of 

-«,  oa,  y  '         9        '  5 

^l?is.  II0?  (a-  — rc^). 

5.  Find  the  least  common  multiple  of 

1  (x  +  ff   49  (x  +  9/)   21a;2  a:  +  .y 

—3 — >  —9—'  -2r'  ^^^^  -i^r- 

Ans.    tm^L±J^^49(x  +  ,yx^. 
o 


6.  Find  the  least  common  multiple  of 
8a-   IBah  X  -f 
3"' "2"' "5 


73a-   7Sah  X  -\-  1/        ^    a 
'     and  — 


Ans.   loa'h  (x  +  ^). 


The  demonstration  being  founded  upon  the  hypothesis,  that  the  frac- 
tions are  reduced  to  their  lowest  terms,  the  rule  is,  of  course,  only 
applicable  to  such  fractions. 


GREATEST  COMMON  DIVISOR  OF  FRACTIONS. 

140.  The  greatest  common  divisor  of  two  or  more  fractions  is  the 
greatest  quantity  that  will  exactlj"^  divide  them,  giving  entire  quotients. 

This  quantity  must  be  a  fraction ;  for  nothing  but  a  fraction  will 
divide  a  fraction  reduced  to  its  lowest  terms,  and  give  an  entire  quo- 
tient. The  greatest  common  divisor  of  several  fractions  will  then  be  a 
fraction  itself;  and  since,  when  we  divide  a  fraction  by  another  frac- 
tion, we  divide  the  numerator  of  the  fraction  assumed  as  the  dividend 
by  the  numerator  of  the  fraction  taken  as  the  divisor,  and  divide  the 
denominator  of  the  divisor  by  the  denominator  of  the  dividend,  it  fol- 
lows, that  the  divisor  sought  must  have  a  numerator  that  will  divide 
each  of  the  numerators  of  the  given  fractions,  and  a  denominator  that 


ALGEBRAIC    FRACTlbXS.  81 

will  be  divisible  by  each  of  the  given  denominators.  But,  since  the 
value  of  a  fraction  increases  with  the  increase  of  its  numerator  and  the 
decrease  of  its  denominator,  it  is  plain  that  a  fraction,  whose  numerator 
would  divide  all  the  given  numerators,  and  whose  denominator  would 
be  divisible  by  all  the  given  denominators,  would  not  be  the  greatest 
fraction  that  will  divide  the  given  fractions,  unless  it  has  the  greatest 
numerator  and  the  least  denominator  that  will  fulfil  the  required 
conditions.  Or,  in  other  words,  unless  its  numerator  is  the  greatest 
common  divisor  of  the  given  numerators,  and  its  denominator  the  least 
common  multiple  of  the  given  denominators. 

RULE. 

141.  Find  the  greatest  common  divisor  of  the  numerators,  and  di- 
vide it  hij  the  least  common  multiple  of  the  denominators  ;  the  fraction 
so  formed  will  he  the  least  common  multiple  of  the  given  fractions. 

EXAMPLES. 

2a'   8a'  4a^  '^^^ 

1.  Find  the  GCD  of  ~,  ~,  J?-,  and  0.(^ 


2.  Find  the  GCD  of 


7.r'    a.^-   (a  +  .r) 


12 '  ;i  ' 


-4-—'  ^"'^  2- 


3.  Find  the  GCD  of  ^,  -y,  *f-„  and  ^. 
d     4    12  db 


4.  Find  the  GCD  ot^,^,—,  and  — . 
za    6     X  X 


5.  Find  the  GCD  of  -^,  -,  ^^~,  and  ^. 

7     21         0  3 

6.  Find  the  GCD  of  —-,  -y,  ^— t— :„  and  -„. 

_x-   X      ax  -f-  X'  X 

Ans. 
Remarks. 


Ans. 

27* 

Ans. 

12" 

A71S. 

1 
3(j' 

1 

ins. 

Qax 

Ans 

X 

I2x'  (a'  +  x) 


142.  The  greatest  common  divisor  of  entire  quantities  may  be  unity, 
and  the  quantities  are  then  said  to  be  prime  with  respect  to  each  other. 
But  it  is  evident  that  fractions  can  never  be  prime  with  respect  to  each 
F 


82  EQUATIONS    OF    THE    FIRST    DEGREE. 

other;  for,  though  their  nuniertators  may  be  prime,  as  in  the  3d,  4th, 
and  6th  examples,  the  least  common  multiple  of  their  denominators  can 
always  be  formed. 

143.  We  may  also  remark  that,  as  the  least  common  multiple  of 
entire  quantities  can  always  be  formed ;  so,  likewise,  all  fractions  what- 
ever have  a  least  common  multiple. 

144.  Knowing  the  least  common  multiple  of  any  number  of  frac- 
tions, we  can  find  their  greatest  common  divisor  by  multiplying  the 
fractious  together,  and  dividing  their  product  by  their  least  common 
multiple. 

145.  Conversely,  knowing  their  greatest  common  divisor,  we  can 
find  their  least  common  multiple  by  multiplying  the  fractions  together, 
and  dividing  the  product  by  the  greatest  common  divisor. 


EQUATIONS  OF  THE  FIRST  DEGREE. 

146.  An  Equation  is  an  expression  containing  two  equal  quantities, 
with  the  sign  of  equality  between  them.  Thus,  cc  =  a  —  &  is  an  equa- 
tion, and  expresses,  that  the  quantity  represented  by  x  is  equal  to  the 
dilTerence  of  the  quantities  represented  by  a  and  b. 

147.  The  part  on  the  left  of  the  sign  of  equality  is  called  the  y?rs^ 
member  of  the  equation,  and  that  on  the  right  the  second  member. 

148.  Problems  can  be  stated  or  expressed,  and  their  solutions  ob- 
tained by  means  of  equations ;  that  is,  we  can  express,  in  algebraic  lan- 
guage, the  relation  between  an  unknown  quantity  to  be  found,  and  one 
or  more  known  quantities,  and,  by  certain  opcratioas,  can  find  the  value 
of  the  unknown  quantity. 

149.  The  expressing  the  relation  is  called  the  statement  of  the  pro- 
blem; and  the  operation  performed  after  the  statement,  to  find  the 
value  of  the  unknown  quantity,  is  called  the  sohUion  of  the  problem. 
Thus,  let  it  bo  required  to  find  a  quantity,  which,  being  added  once  to 
itself,  will  give  a  sum  equal  to  h.  Let  x  be  the  unknown  quantity; 
then,  by  the  conditions,  x  +  x  =  b,  or,  2x  =  b.     Then,  if  twice  x  is 

equal  to  b,  x  itself  must  be  equal  to  half  of  b,  raid  x  =  —  is  the   final 

result.     In  tiiis,  making  the  equation,  .r  -\-  x  =  &  is  tlie  statement, 


EQUATIO-NS    OF    THE    FIRST    DEGREE.  S3 

and  the  subsequent  operation  is  the  solution.  If  the  vakie  found  for 
the  unknown  quantitity  (  —  j  be  substituted  in  the  equation  x  -\-x  =  h, 

we  will  have  —+  —  =/>,  a  true  eqviation.      When  the  value  found 

for  the  unknown  quantity  substituted  in  the  equation  of  the  problem 
makes  the  two  members  equal  to  each  other,  the  equation  is  said  to  be 
satisfied,  and  we  conclude  that  the  solution  is  true. 

150.  The  unknown  quantity,  the  thing  to  be  found,  is  usually  repre- 
sented by  one  of  the  final  letters  of  the  alphabet,  x,  i/,  and  z.  Known 
quantities  are  generally  represented  by  the  first  letters  of  the  alphabet, 
a,  I),  c,  and  cL 

151.  An  equation  with  one  unknown  quantity  is  of  the  first  degree, 
when  the  highest  exponent  of  the  unknown  quantity  in  any  term  is 
unity;  of  the  second,  when  the  highest  exponent  of  the  unknown 
quantity  in  any  term  is  two,  &e. 

X  +  a  =  5  is  an  equation  of  the  first  degree. 

x^  -\-  X  =  a  is  an  equation  of  the  second  degree. 

x'^  -\-  x^  -]-  X  =  a  is  an  equation  of  the  third  degree. 

X*  -\-  px^  +  mx^  +  nx  =  a  is  an  ef[uation  of  the  fourth  degree. 

152.  Equations  which  contain  the  unknown  quantity  from  the  high- 
est to  the  first  power  inclusive,  are  called  complete  equations.  The 
above  are  all  complete  equations  with  one  unknown  quantity. 

153.  Equations  in  which  some  of  the  powers  of  the  unknown  quan- 
tity are  missing,  are  called  incomplete  equations,  x^  =  a  is  an  incom- 
plete equation  of  the  second  degree,  x^  +  x^  =  a  is  an  incomplete 
equation  of  the  third  degree. 

154.  A  numerical  equation  is  one  in  which  all  the  known  quantities 
are  represented  by  numbers.  Thus,  .v^  +  2x  =  4  is  a  numerical 
equation. 

155.  A  literal  equation  is  one  in  which  all  the  quantities,  known 
and  unknown,  are  represented  by  letters.  Thus,  x'^  +  ax  =  h  is  a 
literal  equation. 

156.  An  equation  in  which  the  known  quantities  are  partly  repre- 
sented by  letters,  and  partly  by  numbers,  is  a  mixed  equation.  Thus, 
x^  +  ax  =  h  -\-  2  is  a  mixed  equation. 


84  EQUATIONS  OF  THE  FlUST  DEGREE, 

157.  An  identical  cfjuatiou  is  one  in  wliich  the  two  members  differ 
only  in  form,  if  they  differ  at  aU.  Thus,  2x  =  — ,  ^^^^ —  =  x  +  —, 
and  X  =  X,  arc  identical  equations. 

158.  An  iiiJctcrminatc  equation  is  one  in  which  the  value  of  the 
unknown  quantity  is  indeterminate, 

159.  A  single  equation  with  two  unknown  quantities  is  necessarily 
indeterminate,  for  we  must  assume  the  value  of  one  before  we  can 
determine  that  of  the  other.  Thus,  a:  -f  y  =  10  is  an  indeterminate 
C(]uation,  because,  by  assuming  ?/  =  1,  5,  4,  &,c.,  we  find  x  =  9,  5,  6, 
&c.  And,  by  attributing  an  infinite  number  of  arbitrary  values  to  i/, 
there  will  be  an  infinite  number  of  values  for  x. 

IGO.  All  identical  equations  are  necessarily  indeterminate.  Thus, 
the  equation  x  =  x  will  be  true,  or  satisfied,  when  x  is  1,  10,  1000,  or 
anything  whatever. 

161.  The  oivn  sign  of  a  quantity  is  the  sign  with  which  the  quantity 
is  affected  previous  to  any  operation  being  performed  upon  it.  The 
essential  sign  is  the  sign  with  which  it  is  affected  after  the  operation. 
Thus,  multiply  +  «  by  —  1 ;  the  own  sign  of  a  is  plus,  and  its  essen- 
tial sign  minus.  Eut,  multiply  +  a  by  +  1,  and  the  own  and  essen- 
tial signs  are  the  same.  Subtract  +  ^  from  a,  then  a  —  (-(-&)  =  a 
—  h.     Here,  h  appears  in  the  second  member  with  the  essential  sign. 

162.  Since  a  positive  quantity  cannot  be  equal  to  a  negative,  and, 
conversely,  it  is  evident  that  the  essential  sign  of  the  two  members  of 
an  equation  must  be  the  same. 

163.  Since  quantities  can  only  be  equal  to  quantities  of  the  same 
kind,  it  is  plain  that  the  two  members  of  an  equation  must  be  composed 
of  quantities  of  the  same  kind.  Thus,  if  one  member  represent  time, 
the  other  must  represent  time  also. 

These  two  principles  are  important,  and  ought  to  be  remembered. 

164.  Several  axioms,  or  self-evident  propositions,  are  assumed  as  the 
basis  of  the  principles  by  which  equations  are  solved. 

1.  If  equals  be  added  to  or  subtracted  from  equals,  the  results  will 
be  equal. 

2.  If  equals  be  multiplied  or  divided  by  equals,  the  results  will  be 
equal. 


EQUATIONS    OF    THE    FIRST    DEGREE. 


R'^ 


SOLUTION  OF  EQUATIOXS  OF  THE  FIRST  DEGREE. 

165.  The  transformation  of  an  equation  consists  in  changing  its  form, 
without  destroying  the  equality  of  the  two  members. 
There  are  three  transformations  of  equations. 

First    TraitKjurmatlon. 

IGG.  The  first  transformation  consists  in  clearing  an  equation  of  its 
fractions. 

By  the  second  axiom,  we  have  a  right  to  multiply  botli  nicmhers  of 
an  equation  by  the  same  quantity ;  and  it  is  evident  that  if  we  multi- 
ply the  two  members  by  the  least  common  multij^lc  of  the  denominators, 
the  resulting  equation  will  be  free  of  fractions.  For,  since  each  deno- 
minator will  divide  the  least  common  multiple,  it  is  plain  that  the  pro- 
duct of  each  numerator  by  the  least  common  multiple  will  be  divisible 

by  each  denominator.  Thus,  take  the  equation  —  +  '-+2=4,  mul- 
tiply each  member  by  0,  the  least  common  multiple  of  the  denomina- 
tors, and  the  resulting  equation  is  3x  +  a^  +  12  =  24. 

It  is  plain,  that  if  we  multiply  the  two  members  by  12,  the  product 
of  the  denominators,  the  resulting  equation  will  also  be  free  from  frac- 
tions. For  each  denominator  will  divide  the  product  of  12  by  its 
numerator,  inasmuch  as  it  will  divide  12  itself.  Multiplying  by  12, 
we  get  G.T  +  2ic  +  24  =  48.  Now,  divide  both  members  by  2,  which 
we  have  a  right  to  do  by  the  second  axiom,  and  we  get  3x  -f-  a;  =  12, 
the  same  equation  as  before. 

Again,  take  the  equation  —  +  —  +  ^  +  2  =  5. 

By  the  first  method  (the  least  common  multiple  being  12),  we  get 
i\x  +  2x  +  ox  +  24  =  GO.  By  the  second  method,  we  get  24a;  +  8x 
+  12.r  +  96  =  240.  Divide  both  members  by  4,  and  the  results  are 
the  same.  It  willbe  seen  that  the  second  method  is  the  same  as  mul- 
tiplying each  numerator  into  all  the  denominators  except  its  own,  and 
each  entire  quantity  by  ttie  product  of  all  the  denominators. 

r.ULE. 

Multiplif  hotli  memhcrii  hi/  the  least  common  mulliple,  or  hrj  the  'pro- 
duct of  the  denominators. 
8 


86  EQUATIONS     OF     T II  K     i'lRST     DEGRKE. 

When  tlu^  lonst  coininon  multiple  cau  be  found  without  difficulty, 
the  first  method  is  preferable. 


S'.-cond  Tramfvnnaiion. 

167.  This  consists  in  transposing  known  terms  to  the  second  mem- 
ber, and  unknown  terras  to  the  first  member. 

Take  the  equation  2x  —  2  =  x  -f-  a.  (1) 

By  the  first  axiom,  we  have  a  right  to  add  +  2  to  both  members  of 
the  equation.  Adding,  we  have  2a;  —  2  +  2  =  a;  +  a-r2;  or,  since 
the  —  2  and  +  2  destroy  each  other  in  the  first  member,  the  equation 
becomes  2x  =  x  -\-  a  -\-  2.  By  comparing  this  with  the  equation 
marked  (1),  we  see  that  2  has  passed  into  the  second  member,  by 
changing  its  sign. 

Resume  the  equation  2x  =  x  +  a  +  2,  and  add  minus  x  to  both 
members.  AYe  get  2x  —  x  =  x  —  x  +  a  -{-  2  ;  or,  since  +  x  and 
—  X  destroy  each  other  in  the  second  member,  2x  —  x  =  a  +  2. 
Hence,  x  has  passed  into  the  first  member  by  changing  its  sign. 

We  have  demonstrated  that  quantities  can  be  transposed  from  one 
member  to  another,  provided  we  change  their  signs,  since  transiDosition 
is  nothing  more  than  adding  to  both  members  the  quantities  to  be 
transposed,  with  their  signs  changed.  The  demonstration  can  be  as 
readily  made  by  subtracting  the  quantities  with  their  appropriate  signs. 
Hence,  the  second  transformation  is  equivalent  to  adding  to  both  mem- 
bers the  quantities  to  be  transposed,  with  their  signs  changed,  or  sub- 
tracting these  quantities  with  their  own  signs  from  both  members. 


RULE. 

Change  the  &!<jn  of  each  term  trans/erred  from  one  member  to 
another. 

Third  Tran formation. 

168.  The  third  transformation  consists  in  clearing  the  unknown 
qitantity  of  its  coefficient  or  coefficients. 

Besume  the  equation,  Qx-\-2x  +  ^x  -\  24  =  60;  then,  by  the  second 
transformation,  we  have  6x  -f  2x  -f  Sx  =  60  —  24  =  .36.  This  indi- 
cates that  the  sum  of  6^:",  2x,  and  2>x  is  equal  to  36 ;  hence,  obviously, 
llx  =  36.     Now,  by  the  second  axiom,  we  have  a  right  to  divide  both 


EQUATIONS    OF    THE    FIRST    DEGREE.  87 

members  by  the  same  quantity.  Divide  by  11,  and  we  have  x  =  |f . 
The  unknown  quantity  now  stands  unconnected  with  known  terms,  and 
we  have  its  trae  value. 

Take  ax  -^-hx  —  ex  =■  m. 

By  the   rules  for  factorinir,  we  have  (a  +  h  —  c)  x  =  m;  and  by 

the  second  axiom,  x  =  ; . 

a  -\-  b  —  c 

These  illustrations  show  that  the  third  transformation  is  used  when 
the  equation  has  been  cleared  of  its  fractions,  if  it  contained  any,  and 
when  all  the  known  quantities  in  the  first  member  have  been  removed 
to  the  second,  and  when  all  the  terms  involving  the  unknown  quan- 
tity have  been  placed  in  the  first  member,  if  not  already  there. 

RULE. 

Collect  into  a  single  algebraic  sum  all  the  corfficicnfs  of  the  vnhioicn 
quantity,  and  divide  the  two  memhers  hij  this  sum;  if  there  is  hut  one 
coefficient,  divide  both  members  by  it. 

EXAMPLES. 

1.  Clear  the  unknown  quantity  of  its  coefficient  in  the  equation, 

2.1:  =  6  +  c 

Ans     ,-  =  i±^' 
•> 

2.  Clear  the  unknown  quantity  of  its  coefficient  in  the  equation, 

bx  +  3x  —  ex  =  a. 

Ans.  x  = . 

b  -\-o  —  c 

3.  Clear  the  unknown  quantity  of  its  coefficient  in  the  equation, 

6x  -\-  ax  —  2x  ==  n. 

n 


Ans. 


b  -\-  a 

4.  Clear  the  unknown  quantity  of  its  coefficient  in  the  equation, 

7x  +  3x  +  2x  =  12. 

Ans.  X  =  1. 

5.  Clear  the  unknown  quantity  of  its  coefficient  in  the  equation, 

'2x  —  7x  =  a. 

a  —  a 

Ans.  X  = r  =  — p    . 

—  0  o 


88  EQUATIONS    OF    THE    FIRST    DEGREE. 

1G9.  It  will  be  seen  tliat  the  object  of  the  three  traiisforniatious  is 
to  free  the  unknown  quantity  of  its  connection  with  known  terms,  and 
make  it  stand  alone  in  the  first  member  of  the  equation.  When  so 
situated,  its  true  value  is  known,  being  expressed  by  the  second  mem- 
ber of  the  equation. 

170.  The  various  steps  in  the  solution  of  an  equation,  can  be  best 
illustrated  by  taking  an  example  which  will  require  all  three  trans- 
formations. 

Take  •     |-f  ^+  fl  =  6_2:r. 

First  step.     To  clear  the  equation  of  its  fractions. 

Result.  5x  +  ^x  +  10a  =  10Z>  —  20a:. 

Second  step.  To  get  the  known  terms  to  second  member,  and  un- 
known to  first  member. 

Result  bx  -\-  ^x  +  20a)  =  lOi  —  10a. 

Third  step.     To  clear  the  unknown  quantity  of  its  coefficient. 

Result.  X  =  - — ^ = ~ \ 

171.  Frequently,  only  two  transformations  are  necessary  in  the  solu- 
tion of  an  equation ;  sometimes  only  one,  but  always  at  least  one. 

Take  ax  -\-  h  =  c. 

r  —  h 
By  second  transformation,  ax  ^  c  —  h ;  and  by  third,  x  = . 

Take  2x  —  ox  =  m}  then,  by  third  transformation,  —  x  =  m. 

In  this  the  unknown  quantity  appears  with  the  negative  sign ;  but 
we  were  required  to  find  the  value  of  +  a;  and  not  of  —  x.  Multiply 
both  members  by  minus  unity,  which  we  have  a  right  to  do  by  the 
second  axiom,  and  —  x  becomes  positive.  Hence,  x  =  —  m  is  the 
required  value. 

172.  In  general,  when  the  sign  of  the  unknown  quantity  is  negative 
in  the  final  result,  and  the  conditions  of  the  problem  require  it  to  be 
aflfected  with  the  positive  sign,  both  members  must  be  multiplied  by 
minus  unity. 

173.  We  observe,  also,  that  the  signs  of  all  the  terms  of  an  equation 


EQUATIONS     OF     THE     FIRST     DEGREE.  89 

can  be  changed  without  altering  the  equality  of  the  two  members,  since 
we  have  a  right  to  multiply  these  members  by  minus  unity. 

174.  From  the  foregoing  principles  we  derive,  for  the  solution  of 
equations  of  the  first  degree,  the  following 


RULE. 

I.  If  the  equation  contain  fractions,  dear  it  of  them  hi/  vmltiplying 
hath  members  hi/  the  product  of  the  denominators,  or  their  least  common 
mtdtiple.  ^ 

II.  Bring  all  the  terms  involving  the  unJcnown  quantity/ to  the  first 
memher  {if  not  already  there'),  and  all  the  known  terms  to  the  second. 

III.  Collect  into  a  single  algebraic  sum  all  the  coefficients  of  the  un- 
knotcn  quantity,  and  divide  hoth  members  hi/  this  sum. 

IV.  If  the  nnl-nown  quantiti/  appear  in  the  final  result  icith  a  nega- 
tive sign,  midiiplg  both  members  hi/  minus  unitij. 

EXAMPLES. 

1.  Solve  the  equation  x  +  3.r  —  \x  =  14.  Ans.  x  =  4. 

2.  Solve  the  equation  4x  -f  -J^  —  '—-{-—'  =  50. 

.Ins.  X  =  10. 

3.  Solve  the  equation  ^  +  3.c  +  :^  +  ^  —  GO  =  —  G. 

Ans.  X  =  15. 

4.  Solve  the  equation  2x  -\-  3x  +  ix  —  7  =  —  |. 

Ans.  X  =  J. 

(.    CI  1      ii  .'       ^         X    ,     X         X         X        24a;  176 

5.  Solve  the  equation  _+-+_+  -  -  +  _ _= —. 

J  d  4  0         12  7  < 

Ans.  X  =  12. 

6.  Solve  tho  equation  2x  +  3.x  +  43-  —  7  =  —  4. 

Ans.  X  =  i. 

7.  Solve  the  equation  ~ \- \-  ox  =  —  4-  oc  +  1. 

a         c  a 

Ans.  x  =  c. 


90         EQUATIONS  OF  THE  FIRST  DEGREE. 

8.  Solve  tlie  equation  ».r  -\-  —^  -{■  4x  —  5  =  n''  -f-  4  («-  —  1). 

Ans.  X  =  ?i^ 

X 

9.  Solve  the  equation  — —  -\-  -ix  -{-  h  —  o  =  56  +  3a  -f  1. 

Ans.  x  =  a  -\-  h. 

X  X  X  X 

10.  Solve  the  equation \-  -r-  -\ -,  =  ('Z*rc7. 

a         b         c         (I 

(ahcdf 


Ans.  X  = 


Led  +  acd  -j-  al)d  —  ahc 

11.  Solve  the  equation  -^  —  -r  +  —  +  - =  + 

2         4         ^>         9;t        a 

(P^  -\- pin  -f-  pa  -f  4/^  +  4m  -{-  4«)       p  +  «        T/'  +  m) 

4p  9?l  (? 

Ans.  cc  =  p  -f-  ?n  +  a. 

z 

12.  Solve  the  equation  z  A \-  ab  +  a::  =  m. 

a 

,  a  (in  —  ab) 

Ans.  z  =  --^ -!. 

a'  +  «  +  1 

13.  Solve  the  equation  t/  —  -^  +  ^+  a  +  b^  =  a  -}-  -^  —  c. 

to  o 

—  105c 

^--^-uT+sEb- 

14.  Solve  the  equation  -^^  +  ^  —  ^  +  a  +  3  —  67j  =  10. 

Sab 

(7  4-  GZ.  —  a) 

Ans.  y  =  Ga  '-—- ~^. 

Ida  -f  6 

175.  A  few  examples  follow  of  equations  in  whixih  the  unknown 
quantity  is  affected  with  negative  exi^onents,  and  in  which  the  degree 
of  the  equation  is  apparently  higher  than  the  first. 


2.r-'  -4-  X  —  1        2i'  +  4 
1.  Solve  the  equation -; =  — . 


A71S.  X  =  2. 


2.  Solve  the  equation  p 1-1  =  2. 

ox        "  Ans.  X  =  ^. 

3.  Solve  the  equation  ^  +  -—  =  -^—- :'. 

5  5  5  Ans.  x  =  5. 


EQUATIONS  OF  THE  FIRST  DEGREE.  91 

,     ^  ,        ,  .      ,'  +  7x'  —  Sx        12.;;^  +  14.7;^ 

4.  Solve  the  equation = -: x. 

Ans.  rr  =  6. 
a;-i        X        2x~^  1 

5.  Solve  the  equation  zr-r  +  '—  —  '-^^r—  =  ;; — ;•  .  , 


G.  Solve  the  equation  x~^  -\ h  —  =  , 


1    _    h        a  +  1  +  ah 
ax 


Ans.  .T  =  — . 
o 


7.  Solve  the  equation  1  ■{■  I  +  1  +  o'j  +  i  =  SOx"'. 

Ans.  X  =  40. 

X-*         h  c 

8.  Solve  the  Cfiuatiou  ax~'  ■\ = f-  — ,• 

ax         X 

a  {c  —  a) 

Ans.  X  =  — H T-- 

1  —  ab 

q  1  Q 

9.  Solve  the  equation  2x-^  +  — 2  +  -2  =  ^^~^  +  — 2- 

Ans.  a;  =  1 . 

Gax-^       2m 

10.  Solve  the  equation  2x~^  =  — ^r—  H . 

o  X 

Ans.  0:  =  a  -{■  7n. 

170.  Solufions  of  prohhms  produciuij  equations  of  the  first  di^yree 
with  an  unknoicn  quantity. 

It  has  been  remarked,  that  the  solution  of  a  problem  consist.s  of  two 
distinct  parts,  the  statement  and  the  solution. 

The  examples  just  uiven,  illustrate  the  manner  of  solving  the  equa- 
tion after  the  statement  has  been  made. 

No  general  rules  can  be  given  for  making  the  statement,  which  de- 
pends mainly  upon  the  ingenuity  of  the  student. 

Sometimes  there  is  but  a  single  thing  to  be  determined,  and  in  that 
case,  by  representing  it  by  one  of  the  final  letters  of  the  alphabet,  and 
following  the  conditions  of  the  problem,  we  get  the  equation  or  state- 
ment required. 

Often,  however,  two  or  more  things  are  to  bo  determined,  then  the 
undetermined  quantity,  upon  which  the  other  quantities  depend,  is  re- 
presented by  one  of  the  final  letters  of  the  alphabet. 

As  an  example  of  the  first  class  of  problems,  let  it  be  required  to 
find  a  quantity,  which,  added  to  half  itself,  will  give  a  sum  equal 
to  6. 


92  EQUATIONS  or  Tin:   first  degree. 

Let  X  represent  the  quantity,  then  half  of  the  quantity  will  be  repre- 
sented by  l.r,  and  by  the  conditions  we  liavc  x  -\-  Jx  =  G.  Solving 
the  equation,  we  get  x  =  i. 

As  an  example  of  the  second  class  of  problems,  take  the  following : 
A  man  invests  half  of  his  money  in  State  stocks,  and  a  third  of  the 
remainder  in  a  Savings'  Institution,  and  has  four  hundred  dollars  left. 
Required  the  entire  amount  of  his  capital,  and  the  amount  of  each  of 
his  two  investments. 

Since  his  investments   depend    upon   his  capital,   let  x  =  capital. 


placed  in  Savings'  Institution.  The  remainder,  after  these  investments, 
is  plainly  x  —  (Jx  -f  i^) )  but  this  remainder  is,  by  the  conditions  of 
the  problem,  equal  to  four  hundred  dollars.  Hence,  x  —  (•].«  -|-  \x^ 
=  400.     Solving,  we  get  x  =  1200. 

Then,         i^x  =    600,  Investment  in  State  stock. 

ix  =    200,  "  in  Savings'  Institution. 

400,  The  amount  uninvested. 

1200,  Original  capital. 

The  solution  has  been  verified  by  adding  the  two  investments  to 
what  was  left,  and  getting  a  sum  equal  to  the  original  capital.  The 
solutions  of  all  problems  ought  to  be  verified  in  a  similar  manner. 

We  have  seen  that  an  equation  is  said  to  be  satisfied  when  the  value 
found  for  the  unknown  quantity  substituted  in  the  given  equation, 
makes  the  two  members  equal  to  each  other 

But  the  value  of  the  unknown  quantity  in  the  solution  of  a  problem 
must  not  only  satisfy  the  equation  of  the  problem  (the  statement),  but 
must  also  fulfil  the  required  conditions.  In  consequence  of  the  use  of 
negative  quantities  in  Algebra,  the  equation  of  the  problem  (the  state- 
ment), is  often  satisfied,  when  the  required  conditions  are  not  fulfilled 
in  a  strict,  arithmetical  sense. 

As  an  illustration,  let  it  be  required  to  find  a  quantity,  which,  when 
added  to  4,  will  give  a  sum  equal  to  2.  Let  x  be  the  quantity,  then 
•r  +  4=2.  Hence,  x  =  —2.  Now,  —  2  will  satisfy  the  equation 
of  the  problem,  but  will  not  fulfil  the  conditions  in  an  arithmetical 
sense;  for  we  were  required  to  find  a  quantity,  which,  when  added  to 
4,  would  give  a  sum  equal  to  2,  and  the  quantity  found  is  really  to  be 
taken  from  4. 


EQUATIONS    or    THE    FIRST    DEGREE.  93 

It  will  be  seen  hereafter,  that  there  are  several  cases  in  which  the 
equation  may  be  satisfied,  and  the  conditions  not  fulfilled. 


A  farmer  has  two  kinds  of  oats ;  the  one  kind  worth  80  cents  per 
bushel;  the  other  20  cents  per  bushel.  He  mixes  50  bushels  of  the 
superior  article  with  a  certain  amount  of  the  inferior,  and  sells  the  mix- 
ture at  22  cents  per  bushel.  How  much  of  the  second  quality  did 
he  put  with  the  first  in  order  to  make  the  rate  of  sale  the  same? 

Let  X  represent  the  amount  of  the  inferior  article ;  then,  by  the  con- 
dition, 50  .  30  -f  2Qx  =  (50  -H  x)  22.  Hence,  x-=  200.  In  this 
example,  the  verification  of  the  equation  and  the  fulfilment  of  the  con- 
ditions are  the  same. 

2.  A  farmer  has  50  bushels  of  oats,  worth  30  cents  per  bushel,  and 
200  bushels,  worth  20  cents  per  bushel.  He  mixes  the  two  lots  to- 
gether.    At  what  price  ought  he  to  sell  the  mixture  ? 

Let  X  represent  the  required  price  of  the  mixture  per  bushel;  then, 
50  .  30  -f  20  .  200  =  250x.     Hence,  x  =  22. 

3.  What  number  must  be  subtracted  from  the  numerator  and  deno- 
minator of  the  fraction  J,  to  make  the  new  fraction  the  reciprocal  of 
the  old  ?  Ans.  x  =  7. 

4.  What  number  must  be  subtracted  from  the  numerator  and  denomi- 
nator of  any  fraction  —  to  make  the  new  fraction,  the  reciprocal  of  the 
old.  Ans.  X  =  m  +  n. 

The  solution  is  general,  and  is  true  for  any  fraction  whatever. 

5.  What  number,  added  to  the  numerator  and  denominator  of  the 
fraction  s,  will  make  the  new  fraction  4  times  as  great  as  the  old  ? 

Ans.  X  =  —  3^. 
What  does  this  value  satisfy  ? 

6.  What  number,  added  to  the  numerator  and  denominator  of  an_\ 

fraction  — ,  will  make  the  new  fraction  4  times  as  great  as  the  old  ? 
11  ° 

Ans.  X  = 


•4m 

7.  What  number,  added  to  any  fraction  — ,  will  give  a  sum  equal 

to  the  numerator.  ^  m  (n  —  1) 

Ans.  X  =  — ^^ . 


94  EQUATIONS    OF    THE    FIRST    DEGREE. 

8.  What  number,  added  to  any  fraction  — ,  will  give  a  sum  equal  to 

the  denominator ?  .  n'  — m 

Ans.  X  = . 

n 

In  what  case  will  this  result  fall  to  satisfy  the  conditions   of  the 

problem  ? 

9.  What  number,  multiplied  by  any  fraction  — ,  will  give  a  product 

equal  to  the  sum  of  the  numerator  and  denominator? 

_  (m  4-  n)  n 
m 
How  may  the  last  four  results  be  made  applicable  to  any  fraction 
whatever  ? 

10.  In  a  square  floor  of  a  college  building  there  is  a  certain  number 
of  brick ;  if  one  more  brick  is  added  to  each  side  of  the  floor,  and  the 
square  form  be  preserved,  there  will  be  61  more  brick  than  at  first. 
How  many  brick  does  the  floor  contain?  Ans.  x^  =  900. 

11.  Two  pipes  lead  into  a  reservoir  capable  of  containing  160  hogs- 
heads of  water.  The  first  pipe  can  fill  it  in  16  hours,  and  the  two  to- 
gether in  12  hours.     In  what  time  can  the  second  alone  fill  it  ? 

Ans.  X  =  48  hours. 

12.  Two  travellers  travel  at  the  same  rate,  the  one  starting  before 
the  other  J  at  12  o'clock  the  first  has  travelled  4  times  as  far  as  the 
second,  but  when  they  had  each  gone  15  miles  further,  the  entire  dis- 
tance passed  over  by  the  first  was  only  double  that  of  the  second. 
Eequired  these  distances.  Ans.  45  and  22  J  miles. 

12.  The  sum  of  two  numbers  is  a,  and  their  diiFerence  h.  Required 
the  two  numbers. 

Ans.  The  greater,  7^+7^,  the  smaller  —  —  — . 

AVe  see  from  this  result  that,  knowing  the  sum  and  difference  of  two 
quantities,  we  get  the  greater  by  adding  the  half  sum  to  the  half  diffe- 
rence, and  the  less,  by  subtracting  the  half  difference  from  the  half 
sum. 

This  formula  is  of  extensive  application,  and  ought  to  be  remem- 
bered. 

13.  A  Californian  gold  digger  wishes  to  sell  a  vessel  full  of  gold 
mixed  with  sand.  The  vessel,  when  filled  with  gold,  will  weigh  ten 
pounds,  and  when  filled  with  sand,  one  pound;  the  mixture  weighs  seven 
pounds.     How  much  gold  is  in  it  ?  Ans.  Q-^\  lbs. 


GEOMETRICAL     PROPORTION.  95 

14.  The  railing  around  the  altar  of  the  Cathedral  in  the  City  of 
Mexico  is  a  composition  of  gold  and  silver.  Assuming  that  279  pounds 
of  the  composition  loses  20  pounds  when  immersed  in  water,  and  that 
19  i  pounds  of  gold  lose  one  pound  in  water,  and  10^  pounds  of  silver 
lose  one  pound  in  water.  Required  the  proportion  of  gold  and  silver 
in  the  alloy.  Ans.  Gold  :  Silver  :  :  156  :  123. 

15.  Two  men  purchase  together  a  barrel  of  flour,  weighing  196 
pounds,  for  5  dollars ;  the  first  pays  half  a  dollar  more  than  the  second. 
How  ought  they  to  divide  the  flour  ? 

.4ns.  The  first  oudit  to  have  1074  lbs,  the  second  881. 


GEOMETRICAL    PROPORTION. 

177.  Ratio  is  the  quotient  arising  from  dividing  one  quantity  by 
another  of  the  same  kind.      Thus,  if  M  and  N  represent  quantities 

of  the  same  kind,  then  —  expresses  the  ratio  of  M  to  X. 

Four  quantities,  M,  N,  P,  and  Q,  are  said  to  be  proportional  when 
the  ratio  of  the  first  to  the  second  is  the  same  as  that  of  the  third  to 
the  fourth. 

Thus,  04,  8,  10,  and  2  constitute  a  proportion,  because  the  first, 
divided  by  the  second,  is  equal  to  the  third  divided  by  the  fourth. 

The  proportionality  of  four  quantities,  M,  N,  P,  and  Q,  is  expressed 
thus,  M  :  N  :  :  P  :  Q,  and  is  read,  M  is  to  N  as  P  is  to  Q. 

The  first  and  last  terms  of  a  proportion  are  called  the  extremes  ;  the 
second  and  third  the  means.  Of  four  proportional  quantities,  the  first 
and  third  are  called  antecedents  ;  the  other  two,  consequents.  Of  three 
quantities,  M,  N,  and  P ;  when  M  :  X  :  :  N  :  P,  then  X  is  said  to  be 
a  mean  proportional  between  M  and  P;  and  X  is,  at  the  same  time, 
antecedent  and  consequent. 

Two  quantities,  M  and  X,  are  said   to  be  reciprocally  proportional, 
when  the  one  increases  as  fast  as  the  other  diminishes.     One  of  the 
quantities  must  be  equal  to  a  fraction  with  a  constant  numerator,  and 
with  the  other  quantity  for  its  denominator. 
4 
X 
tional. 


96  Ci  K  V  M  i;  T  l\  I  C  A  J.     1'  R  0  P  0  R  T  I  0  N  . 

Equi-multiples  of  two  quantities  are  the  products  wliicli  arise  from 
multiplying  tlicm  by  the  same  quantity.  Thus,  aJI  and  oN  are  equi- 
multiples of  M  and  N.  the  common  factor  being  a. 

Theorem  I. 

If  four  quantities  are  in  proportion,  the  product  of  the  extremes  ■will 
equal  the  product  of  the  means. 

]M       P 

Foi',  when  M  :  N  :  :  P  :  Q,  we  know  that  —  =  — ;  and  by  clearing 

of  fractions,  MQ  =  PN.. 

This  theorem  furnishes  an  important  test  of  the  proportionality  of 
four  quantities.  Whenever  the  product  of  the  extremes  is  equal  to 
that  of  the  means,  the  proportion  is  true ;  and  when  that  is  not  the 
case,  it  is  a  false  proportion. 


Theorem  II, 

178.  When  the  product  of  two  quantities  is  equal  to  the  product  of 
two  other  quantities,  two  of  them  may  be  taken  as  the  extremes,  and 
two  as  the  means  of  a  proportion. 

For,  suppose  ]MQ  =  NP,  divide  both  members  by  QN,  the  first  mem- 

M  P  MP 

ber  will  become  — ,  and  the  second  — ,  and  we  have  —  =  — .       From 

IN  \^  li  Vc^ 

which  we  get  the  proportion  INI  :  N  :  :  P  :  Q. 
Let  2  .  4  =  8  .  1.     Then  2  :  8  :  :  1  :  4. 


Theorem  III. 

179.  The  square  of  a  mean  proportional  is  equal  to  the  product  of 
the  other  two  terms  of  the  proportion. 

For,  from  the  definition  of  a  mean  proportional,  we  have  M  :  N  :  : 
N  :  P ;  hence,  by  Theorem  I.,  N^"  =  MP. 
Let  2  :  4  :  :  4  :  8.     Then  4^  =  2  .  8  =  16. 

Theorem  IV. 

180.  W^hen  four  quantities  are  in  proportion,  they  will  also  be  in 
proportion  by  alternation  ;  that  is,  when  antecedent  is  compared  with 
antecedent,  and  consequent  with  consequent. 

For,   let   M  :  N  :  :  P  :  Q;   then,  by  Theorem  I.,  MQ  =  NP,  and 


GEOMETRICAL  PROPORTION.  97 

dividing  tliis  equation,  member  b}'  member,  by  PQ,  we  get  1^  =  j^ , 

from  whicb  M  :  P  :  :  N  :  Q. 

Let  4  :  8  :  :  12  :  24.     Then,  also,  4  :  12  :  :  8  :  24. 


Theoiiem  V. 

181.  When  four  quantities  arc  in  proportion,  they  will  be  in  propor- 
tion when  taken  inversely/;  that  is,  when  the  consec^uents  take  the 
place  of  antecedents,  and  tlic  antecedents  the  place  of  consequents. 

Let  M  :  N  :  :  P  :  Q;  then,  also,  we  have  MQ  =  XP,  or  NP  =  MQj 

divide  both  members  by  ^MP,  and  we  get  ^  =  — ,  from  which  X  : 
M  :  :  Q  :  P. 

Theorem  VI. 

182.  If  four  quantities  are  in  proportion,  they  will  be  in  proportion 
by  composition  ;  that  is  when  the  sum  of  antecedent  and  consequent  is 
compared  with  either  antecedent  or  consequent. 

M      P 

Let  M  :  N  :  :  P  :  Q;  thcn^  also,  ^  =  — .    Add  1  to  both  membei-s, 

,         ,  M  +  X        P  +Q 

and  reduce  to  a  common  denominator,  and  we  have  — r^: —  =  — --r — , 

from  which  we  get  jM  +  X  :  N  :  :  P  +  Q  :  Q. 

Let  4  :  12  :  :  8  :  24.     Then  4  -f  12  :  12  :  8  +  24  :  24. 

In  like  manner  it  may  be  shown,  that  when  M  :  N  :  :  P  :  Q,  that 
we  will  also  have  M  +  X  :  :M  :  :  P  +  Q  :  P- 


Theorem  VII. 

183.  When  four  quantities  are  in  proportion,  they  will  also  be  in 
proportion  by  division  ;  that  is,  when  the  difference  of  antecedent  and 
consequent  is  compared  with  either  antecedent  or  consequent. 

M       P 

Let  !iM  :  X  :  :  P  :  Q ;  then,  also,  we  have  —  =  — .     Subtract  1  from 

X  Ki 

both  members,  and  reduce  to  a  common  denominator,  we  will  have 

M  —  N      P  —  Q 

j^       =  —T^,  from  which  we  get  31  —  X  :  X  :  :  P  —  Q  :  Q. 

Let  12  :  4  :  :  24  :  8.     Then  12  —  4  :  4  :  :  24  —  8  :  8. 

It  may  be  shown  in  like  manner,  that  when  four  quantities  are  in 
7  o 


98  GEOMETRICAL     PROPORTION. 

proportion,  M  :  N  :  :  P  :  Q,  tluxt  we  will   also  liave  M  —  N  :  M 
P— Q:P. 

Let  12  :  4  :  :  24  :  8.     Then,  also,  12  —  4  :  12  ::  24  —  8  :  24. 


Theorem  VIII. 

184.  Equi-multiples  of  any  two  quantities  are  proportional  to  the 
quantities  themselves. 

For,  take  the  identical  proportion,  M  :  N  :  :  IM  :  N ;  then  MN  = 
NM.  Multiply  both  members  by  in,  and  there  results  mM  .  N  = 
mN  .  M ;  and,  since  mM  and  ?>iN  may  be  regarded  as  single  terms,  we 
have  from  Theorem  2d,  M  :  N  :  :  mM  :  mN. 

Let  1  and  2  be  multiplied  by  the  same  number,  3 ;  then  1.2:: 
1.3:2.3:  or  1  .  2  :  :  3  :  6. 


Theorem  IX. 

185.  If  equi-multiples  of  the  antecedents  of  four  proportional  quan- 
tities, and  also  equi-multiples  of  the  consequents  be  taken,  the  four 
resulting  quantities  will  be  proportional. 

For,  let  M  :  N  :  :  P  :  Q ;  then  MQ  =  NP.  Multiplying  both  mem- 
bers by  inn,  and  there  results  mM  .  «Q  =  «N  .  5»P.  Hence,  mM  : 
tcN  :  :  mP  :  iiQ. 

Let  1  :  2  :  :  4  :  8,  and  take  m  =  4,  and  n  —  3. 

Then  4  .  1  :  3  .  2  :  :  4  .  4  :  3  .  8 ;  or  4  :  6  :  :  16  :  24. 

It  is  plain  that  the  above  theorem  is  equally  true  when  m  =  n.  So, 
that  all  the  terms  of  a  proportion  may  be  multiplied  by  the  same  quan- 
tity without  destroying  the  proportionality  of  the  terms.  An  infinite 
number  of  proportions  may  then  be  formed  from  a  single  proportion. 

Theorem  X. 

186.  If  there  be  two  sets  of  four  proportional  quantities,  having  two 
terms,  the  same  in  both,  the  remaining  terms  will  constitute  a  pro- 
portion. 

Let  M  :  N  :  :  P  :  Q. 

M  :  N  :  :  R  :  S. 
Then,  MQ  =  NP,  or  NP  =  ^^IQ. 

MS  =  NR  and  NR  =  MS. 


GEOMETRICAL     PROPORTIOX.  99 

P       Q 

Dividing  these  equations  member  by  memocr,  we  get  — -  =  — -,  from 

which,  P  :  R  :  :  Q  :  S,  or  P  :  Q  :  :  R  :  S.     (Art.  180). 

And  the  same  property  can  evidently  be  shown  when  any  other  two 
terms  are  equal. 
Let  1  :  2  :  :  4  :  8. 

1  :  2  :  :  6  :  12.     Then  4  :  G  :  :  8  :  12,  or  4  :  8  :  :  (i  :  12. 
Let  1  :  2  :  :  4  :  8. 

1  :  3  :  :  4  :  12.     Then  2  :  3  :  :  8  :  12,  or  2  :  8  :  :  8  :  12. 

It  will  be  seen  that  the  corresponding  terms  of  the  proportion  must 
be  taken  in  connection  with  each  other. 


Theorem  XI. 

187.  Of  four  proportional  quantities,  if  the  two  antecedents  be  aug- 
mented or  diminished  by  quantities  which  are  proportional  to  the  two 
consequents,  the  resulting  quantities  will  be  proportional  to  the  con- 
sequents. 

For,  let  M  :  N  :  :  P  :  Q. 

N  :  Q  :  :  R  :  S. 
Then,  MQ  =  NP.  (A). 

NS  =  QR. 
Or,  QR  =  NS.  (B). 

Adding  and  subtracting  (B)  from  (A),  there  results  Q  (M  =b  R)  = 
N  (P  d=  S).     Hence,  31  ±  R  :  P  ±  S  :  N  :  Q. 
Let  1  :  2  :  :  4  :  8. 
And  3  :  6  :  :  4  :  8.     Then  1  =b  3  :  2  d=  6  :  :  4  :  8. 

It  can  be  shown  in  like  manner,  that  if  the  two  consequents  be  aug- 
mented or  diminished  by  quantities,  which  are  proportional  to  the 
antecedents,  the  resulting  quantities  Avill  be  proportional  to  the  ante- 
cedents. 

The  reciprocal  of  a  ([uantity  is  unity  divided  by  the  quantity ;  thus, 

the  reciprocal  of  A  is  -r- . 
A 

Theorem  XII. 

188.  Two  quantities  are  inversely  proportional  to  their  reciprocals. 
For,  take  the  identical  proportion  A  :  B  :  :  A  :  B ;  then  AB  =  B  A. 


100  GEOMETRICAL     PROPORTION. 

Divide  botli  members  by  AB ;  then 

Hence,  A  :  B  :  :  -  :  -.     Thus,  2  :  3  :  :  -J  :  ^. 
j>      A 


Theorem  XIII. 

189.  If  there  are  any  number  of  proportions  having  the  same  ratio, 
any  one  antecedent  will  be  to  its  consequent  as  the  sum  of  all  the  ante- 
cedents to  the  sum  of  all  the  consequents. 

For,  let  M:N::P:Q. 

M  :  N  :  :  A  :  B. 
M  :  N  :  :  C  :  D. 
M  :  N  :  :  E  :  F.  J 

Tlien,  MQ  =  NP. 

MB  =  NA. 
MD  =  NC. 
MF  =  NE. 

Adding  these  equations  member  by  member,  we  get  M  (Q  +  -Tj  + 
D  +  F)  =  N  (P  +  A  +  C  +  E). 

Hence,  M:N::P  +  A  +  C+E:Q  +  B  +  D-fF. 


Let  1:2 

1  :2 
1  :2 


Then,  1  :  2  :  :  4  -f  6  +  8  :  8  +  12  +  IC),  or  1  .  2  :  :  18  :  30. 

It  is  not  necessary  that  the  first  antecedent  and  consequent  of  each 
proportion  should  be  M  and  N;  all  that  is  required  is,  that  they  have 
the  same  ratio ;  for,  then,  we  can  substitute  SI  and  N  for  them,  and 
the  above  demonstration  becomes  applicable.  To  show  our  authority 
for  this  substitution,  take  the  two  proportions, 

M  :  N  :  :  P  :  Q,  and  E  :  S  :  T  :  U. 

II       T 

From  the  second,  we  get  RU  =  ST,  or  —  =  — ;  and,  since,  by  hy- 

Pi       M  M      T 

pothesis,  TT  =  ^,  there  results  —  =  — .     Hence,  M  :  N  :  :  T  :  U. 


GEOMETRICAL  PROPORTION.  101 

When  we  have  any  number  of  proportions,  like  those  marlied  A, 
having  a  common  ratio,  we  can  obviously  form  a  continued  proportion 
from  them. 

Thus,         M  :  N  :  :  P  :  Q  :  :  A  :  B  :  :  C  :  D  :  :  E  :  F. 

So,  also,  1  :  2  :  :  4  :  8  :  :  G  :  12  :  :  8  :  16. 

The  character  :  :  is  written  before  each  new  antecedent,  and  refers 
it  and  its  consequent  back  to  the  first  antecedent  and  consequent. 

It  is  obvious  that  any  antecedent  may  be  repeated  any  number  of 
times,  provided,  that  its  consequent  is  repeated  the  same  number  of 
times ;  and,  also,  that  any  consequent  may  be  multiplied  or  divided  by 
HDything  whatever,  provided,  that  the  same  operation  be  performed  on 
its  consequent. 

Thus,  M  :  N  :  :  P  +  A  +  0  +  P:  :  Q  +  B  -f  T)  +  F, 

may  be  written, 

M  :  N  :  :  P  +  A  +  A  +  C  +  mE  :  Q  +  B  +  B  +  D  +  m¥, 
without  altering  the  truth  of  the  proposition. 

Thus,  let  4  :  8  :  6  :  12,  and  4  :  8  :  :  5  :  10. 

Then,  4  :  8  :  :  0  +  5  :  12  +  10,  and  4  :  8  :  :  6  +  5  +  o  :  12  +  10  +  10. 

And  also,  4  :  8  :  :  6  X  0  +  5  +  5  :  12  X  9  +  Id  +  10. 

Theorem  XIV. 

190.  If  four  quantities  are  in  proportion,  the  sum  of  the  first  ante- 
cedent and  consequent  will  be  to  their  difference  as  the  sum  of  the 
second  antecedent  and  consequent  is  to  their  difference. 

For,  since  M  :  N  :  :  P  :  Q,  it  follows  that  MQ  =  NP. 

Add  NQ  to  both  members,  then  Q  (M  +  N)  =  X  (P  +  Q). 

Subtract  XQ  from  both  members,  then  X  (P  —  Q)  =  Ql  —  X)  Q. 

Multiplying  the  two  equations  together,  there  results 

(M  +  X)  (P  —  Q)  :=.  (P  +  Q)  (M  -  X). 
From  which  we  get  M  +  X^  :  M  —  X  :  :  P  +  Q  :  P  —  Q. 
Let  2  :  6  :  :  8  :  24. 

Then  2  +  6  :  2  —  6  :  :  8  +  24  :  8  —  24,  or  8  :  —  4  :  :  32  :  — 16. 
9* 


102  Ci  EOMETRICAL    PROPORTION 


Theorem  XV. 


191.  If  there  are  any  number  of  proportional  quantities,  a  single 
proportion  may  be  formed  from  them  by  multiplying  together  the  cor- 
responding terms  of  all  the  proportions. 

For,  let  M  :  N  :  :  P  :  Q. 

A  :  B  :  :  C  :  D. 
E  :  F  :  :  a  :  H. 

Then,  MQ  =  NP. 

AD  =  BC. 
EH  =  FG. 

Multiplying  these  equations,  member  by  member,  we  get 

(MAE)  (QDH)  =  (NBF)  (PCG). 

Hence,  MAE  :  NBF  :  :  PCG  :  QDH. 

Let  1  :  2  :  :  4  :  8. 

3  :  1  :  :  9  :  3. 
5  :  10  :  :  2  :  4. 

Then,  also,  15  :  20  :  :  72  :  9G. 

Theorem  XVI. 

192.  If  four  quantities  are  in  proportion,  their  like  powers  will  be 
proportional. 

For,  let  M  :  N  :  :  P  :  Q. 

Then,  MQ  =  NP. 

Squaring  both  members,    M'^Q^  =  N'P'. 

Hence,  M^  :  N^  :  :  P^  :  Ql 

In  like  manner  it  may  be  shown  that  M'"  :  N"  :  :  P™  :  Q"". 

Let  1  :  2  :  :  4  :  8. 

Then,  also,       1-  :  2^  :  :  4"  :  8^  or  1  :  4  :  :  16  :  04. 

Likewise,         1=  :  2^  :  :  4^  :  8^  or  1  :  8  :  :  64  :  51  2. 

And  similar  proportions  can  be  obtained  for  the  4th,  5th,  &c., 
powers. 


GEOMETRICAL    PROPORTION.  103 


General  Remarlcs. 

193.  All  the  foregoing  theorems  have  been  deduced  from  the  first 
two ;  and  it  is  obvious  that  an  infinite  series  of  proportions  might  be 
deduced  from  them.  The  test  of  the  truth  of  any  proportion  deduced 
directly  or  indirectly  from  the  first  two  theorems,  is  the  product  of  the 
extremes  being  equal  to  the  product  of  the  means.     Thus,  if 

M  :  N  :  :  P  :  Q;  then,  also,  M  +  4  :  P  +  4  ^  :  N  :  Q. 

For,  by  multiplying  tlie  extremes  and  means  together,  we  have 

]MQ  -f  4Q  =  NP  +  4Q, 

which  is  a  true  equation  when  MQ  =  NP.  In  general,  no  propor- 
tion is  false,  however  absurd  it  may  seem,  when  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means.  Thus,  \  :  x*  :  :  x~*  :  1 
is  a  true  proportion,  because  the  pi-oduct  of  the  extremes  is  equal  to 
unity,  and  that  of  the  means  also  equal  to  unity. 

The  proportion  M  :  N  :  :  P  :  Q,  gives  MQ  =  NP,  from  which  we 
can  get  the  four  equations 


M 

P 

N 

~Q' 

i\r 

N 

1^ 

~Q' 

Q 

N 

~M' 

Q 

N 

P 

~M' 

involving  eight  distinct  ratios.  And,  since  ratio  is  the  quotient  arising 
from  dividing  one  quantity  by  another  of  the  same  kind,  we  conclude, 
from  the  above  series  of  equations,  that  either  the  first  consequent  or 
the  second  antecedent  must  represent  quantities  of  the  same  kind  as  the 
first  antecedent;  and  that  either  the  first  antecedent  or  the  second 
consequent  must  represent  quantities  of  the  same  kind  as  the  first  con- 
sequent. We  also  see  that  two  distinct  species  of  quantities,  and  but 
two,  can  be  represented  by  a  proportion.  If  two  terms  of  a  proportion 
are  abstract  numbers,  the  proportion  can  represent  but  one  kind  of 
quantity. 


104  GEOMKTRICAL     PROPORTION. 

104.  In  tlie  equation  MQ  =  NP,  resulting  from  the  proportion 
31  :  N  :  :  P  :  Q,  if  three  terms  are  known,  it  is  plain  that  the  fourth 
term  can  be  found  by  solving  the  equation  with  reference  to  it.  In 
the  Single  Rule  of  Three  of  Arithmetic,  three  terms  are  given  to  find 
the  fourth.  Thus,  suppose  the  first  three  terms  of  a  proportion  are  3, 
4,  and  G.  The  fourth  term  can  be  found  from  the  proportion  3:4:: 
6  :  ,r.  Hence  3.r  =  24,  or  x  =  S,  and  the  complete  proportion  is 
3  :  4  :  :  6  :  8.  So,  likewise,  let  the  first  three  terms  be  a,  b,  and  c, 
then  the  fourth  results  from  the  proportion,  a  :  h  :  :  c  :  x.     Hence, 

.(■  =  — .  And  we  see  that  the  fourth  term  can  be  found  by  multiply- 
ing the  second  and  third  together,  and  dividing  their  product  by  the 
first. 

If  the  first,  or  either  of  the  middle  terms  is  unknown,  we  have  only 
to  represent  the  unknown  term  by  x,  form  the  equation  from  the  pro- 
portion, and  then  solve  it  with  reference  to  x. 

Let  the  last  three  terms  of  a  proportion  be  a,  h,  and  c,  to  find  the 

first.     We  have,  then,  x  :  a  :  :  b  :  c.     Hence,  x  =  — . 

c 

So,  we  see  that  either  extreme  can  be  found  by  multiplying  together 

the  means,  and  dividing  their  product  by  the  other  extreme. 

Let  the  second  term  be  unknown,  and  the  other  three  be  a,  b,  and  c. 

Then  a  :  x  :  :  b  :  c.  Hence,  a;  =  y.  Let  the  third  term  be  unknown, 
and  the  other  three  be  a,  b,  and  c.  Then  a  :  b  :  :  x  :  c.  Hence,  x  .— 
— .  We  see  that  either  mean  may  be  found  by  multiplying  the  ex- 
tremes together,  and  dividing  their  product  by  the  other  mean. 


EXAMPLES. 

1.  The  first  term  of  a  proportion  is  a^;  the  third,  c^;  the  fourth,  ml 
What  is  the  second  term  ? 

Ans.   — ~. 
c 

2.  The  first  term  of  a  proportion  is  o">+"5  the  second,  a-° ;  the 
fourth,  o""".     AVhat  is  the  third  term  ?  Ans.   a'". 

3.  The  three  last  terms  of  a  proportion  are  a"",  £%  and  (aby^    What 
is  the  first  term  ?  Ans.   a~^b~\ 


I  ox.  105 

4.  Given  the  first  three  ternit;  of  a  prcportiou,  a^  —  W^  a  -\-  b,  and 
a  —  b,  to  find  the  fourth  term.  A7is.  1. 

5.  Given  three  finst  terms  of  a  proportion,  a',  i"",  and  c"",  to  find  the 
fourth.  \  Ans.  b''c'^a~'. 


PROBLEMS   IX   GEOMETRICAL  PROPORTIOX. 

195.  1.  Two  numbers  are  to  each  other  as  2  to  3;  but,  if  8  be 
added  to  both,  the  sums  thence  arising  will  be  to  each  other  as  4  to  5. 
What  are  the  numbers  ?  Ans.  8  and  12. 

Let  X  =  smaller,  then  the  greater  will  be  %x.  For,  2  :  3  :  :  a-  to 
fourth  term  |x.     Then  .r  +  8  :  -|a;  +  8  :  :  4  :  5. 

2.  An  author  can  write  240  pages  in  ten  weeks  :  how  many  can  he 
write  ii\12J  weeks?  Ans.  300  pages. 

3.  Two  quantities  are  to  each  other  as  a  is  to  J;  but,  if  c  be  added 
to  both  of  them,  the  resulting  sums  will  be  to  each  other  as  d  to  c. 
What  are  the  quantities  ? 

.  (e  —  J)  or        ,  (e  —  d)  be 

Ans.   ~- —  and  \-^ — . 

ua  —  ae  bd — ae 

4.  An  author  can  write  a  pages  in  h  weeks :  suppose  that  he  writes 
uniformly,  how  many  pages  can  he  write  in  c  weeks  ? 

Ans.  X  =  —  pages. 

What  values  must  be  given  to  the  letters  in  order  to  make  Examples 
3  and  4,  the  same  as  1  and  2  ?  How  can  the  results  in  the  last  two 
examples  be  verified  ? 

5.  A  father's  age  is  now  3  times  that  of  his  son,  but  in  10  years 
more  the  father's  age  will  only  be  double  that  of  the  son.  What  are 
their  respective  ages  now  ?  Ans.  Father,  30;  son,  10. 

6.  A  father's  age  is  now  a  times  greater  than  that  of  his  son ;  but 
in  b  more  years  the  father  will  only  be  c  times  older  than  his  son. 
What  arc  the  respective  ages  of  father  and  son  ? 

Ans.   Father,  ^^ ^  ;    son, — . 

a  —  c  a  —  c 

7.  A  gcntl,^:;i;ui  puts  out  liis  money  at  a  certain  rate  of  interest,  and 
derives  $500  income.  He  puts  the  same  sum  out  a  second  year,  one 
per  cent,  more  advantageously,  and  derives  6600  from  it.  What  are 
the  two  rates  of  interest  ?  A7is.  5  and  6  per  cent. 


106  GEOMETRICAL    PROPORTION. 

8.  The  governor  of  a  besieged  town  has  provisions  enougli  to  allow 
each  of  the  garrison  1*  pounds  of  bread  per  day,  for  GO  days;  but,  as 
he  learns  that  succor  cannot  be  expected  for  80  days,  he  finds  it  neces- 
sary to  diminish  the  allowance.     What  must  the  allowance  be  ? 

Ans.   IJ  lbs.  per  day. 

9.  A  gentleman,  who  owns  20  slaves,  had  laid  in  a  twelve-months' 
supply  for  them,  when  he  purchased  10  more.  How  long  will  his  sup- 
plies last?  A)is.  8  months. 

10.  A  planter,  who  knows  that  his  negro-man  can  do  a  piece  of  work 
in  6  days,  when  the  days  are  12  hours  long,  asks  how  long  it  will  take 
him  when  the  days  are  15  hours  long.  Ans.  4  days. 

11.  Three  negroes  can  hoe  a  field  of  cotton  in  7  days  :  how  long  will 
it  take  4  to  do  the  same  work?  Ans.  51  days. 

12.  A  family  of  5  persons  use  a  barrel  of  flour  in  C  weeks :  how 
long  will  2  barrels  last  7  persons,  using  it  at  the  same  rate. 

Ans.  8 1  weeks. 

13.  Two  farmers  purchase  a  piece  of  land  for  $1000;  whereof  the 
first  pays  $600,  and  the  second  $400,  and  sell  it  again  for  $1200. 
What  proportion  of  the  pi'ofit  ought  each  to  receive  ? 

Ans.  The  first,  $120;  the  second,  $80. 

14.  A  bookseller  purchased  a  certain  number  of  books  at  the  rate  of 
2  for  a  dollar,  and  also  an  equal  number  at  the  rate  of  3  for  a  dollar, 
and  sold  the  whole  at  the  rate  of  5  for  3  dollars,  and  gained  $22  by 
the  sale.     What  was  the  entire  number  of  books  he  bought  and  sold  ? 

A71S.  120. 

15.  The  hour  and  minute  hands  of  a  clock  are  together  at  12  o'clock. 
When  will  they  be  together  again  ? 

Ans.  5j^Y  minutes  past  one  o'clock. 
Let  X  be  the  space  after  one  passed  over  by  the  hour  hand  before 
being  overtaken  by  the  minute  hand.     Then,  12  :  1  :  :  60  -f-  a-  :  .x. 

16.  Two  pedestrians  start  from  the  same  point  at  the  same  time,  to 
walk  around  a  race-course  a  mile  in  circumference.  The  first  walks 
11  yards  in  a  minute,  and  the  second  34  yards  in  3  minutes.  How 
many  times  will  the  first  have  gone  around  the  track  before  he  is  over- 
taken by  the  second  ?  Ans.  33  times. 

17.  A  brick-mason  has  brick  9  inches  long,  and  4^}  inches  wide, 
with  which  to  build  a  wall  112^1  feet  long.  He  wishes  to  place  180 
bricks  in  a  row,  and  to  lay  some  of  them  with  their  ends  to  the  front 


NEGATIVE     QUANTITIES.  107 

as  headers,  and  some  of  them  lengthwise  as  stretchei-s.     How  many 
headers  and  how  many  stretchers  must  there  be  in  each  row? 

Ans.  60  headers,  and  120  stretchers. 

18.  A  brick-mason  wislies  to  place  twice  as  many  stretchers  as 
headers  in  a  wall  105  feet  long.  How  many  of  each  kind  must  he  use, 
supposing  the  dimensions  of  the  brick  the  same  as  in  the  last  problem  ? 

Ans.  112  stretchers,  and  56  headers. 

19.  A  Freshman  recited  5  times  a  week  in  mathematics,  and  his 
average  for  the  week  was  66.  His  average  for  the  first  three  days  was 
to  the  average  for  the  last  two  as  7  to  6.     What  were  those  averages? 

A)u.  70  and  60. 

20.  Three  farmers  purchase  900  acres  of  land  for  S9000 ;  of  which 
the  first  pays  $2000,  the  second  83000,  and  the  third  S4000.  What 
share  ought  each  to  get,  supposing  that  the  land  is  equally  valuable 
throughout  ? 

Ans.  The  first,  200  acres ;  the  second,  300,  and  the  third,  400. 

21.  A  composition  of  copper  and  tin,  containing  50  cubic  inches, 
was  found  to  weigh  220  7  ounces.  Assuming  that  a  cubic  inch  of 
copper  weighs  4-66  ounces,  and  that  a  cubic  inch  of  tin  weighs  3-84 
ounces,  what  must  have  been  the  relative  proportion  of  tin  and  copper 
in  the  composition?  Ans.  Tin  to  the  copper  as  1  to  2i. 


NEGATIVE    QUxVNTITIES. 

19G.  Quantities  are  con.sidered  as  negative  Avhcn  opposed,  in  charac- 
ter or  direction,  to  other  quantities  of  the  same  kind  that  are  assumed 
to  be  positive. 

If  a  ship,  sailing  at  the  rate  of  10  miles  per  hour,  encounter  a  head- 
wind that  drives  it  back  at  the  rate  of  8  miles  per  hour,  then  its  rate 
of  advance  will  plainly  be  expressed  by  10  —  8  miles  per  hour.  Here 
the  retrograde  movement,  as  opposed  to  the  forward,  is  considered 
negative.  Suppose  the  wind  to  increase  in  violence  until  the  ship  is 
carried  back  at  the  rate  of  12  miles  per  hour.  Then  its  rate  of  advance 
will  be  expressed  by  10  — ■  12,  or  —  2  miles  per  hour.  The  ship  will 
then  plainly  be  carried  back  at  the  rate  of  2  miles  per  hour,  and  we 
see  that  the  minus  sicn  has  indicated  a  chamre  of  direction. 


108  NKGATIVE     QUANTITIES. 

If  a  man  be  now  thirty  years  old.  His  age,  four  years  hence,  will 
be  expressed  by  30  +  4.  Four  years  ago,  it  would  have  been  ex- 
pressed by  30  —  4.  Here,  future  time  being  positive,  past  time  is  ne- 
gative. And  we  sec  that  a  change  of  character  is  again  indicated  by 
a  change  of  sign. 

If  a  man's  money  and  estate  be  represented  by  a,  and  his  debts  by 
h;  then  a  —  h  will  express  what  he  is  worth.  His  debts,  as  opposed 
to  his  property,  are  considered  negative.  If  h  be  greater  than  a,  then 
a  —  Z>  is  negative ;  and  we  commonly  say  that  the  man  is  worth  less 
than  nothing.  We  see  that  there  has  been  a  change  of  sign  in  the 
expression  for  what  the  man  was  worth,  corresponding  to  a  change  in 
the  estimation  of  that  worth.  When  the  worth  fell  below  zero,  its  ex- 
pression became  negative,  because  regarded  as  positive  when  above 
zero. 

If  a  man  agree  to  labor  for  two  dollars  a  day,  and  to  forfeit  one 
dollar  for  every  day  that  he  is  idle ;  and  he  labor  4  days,  and  is  idle  2 ; 
then  his  wages  will  ^e  4  x  2  +  2  (—  1),  or  4  x  2  —  2  ( -f- 1)  =  8  — 2 
dollars.  We  see  that  we  have  regarded  as  negative  either  the  for- 
feiture, as  opposed  to  the  gain,  or  the  working  days  as  opposed  to  the 
idle. 

Distance,  regarded  as  positive,  when  estimated  in  one  direction,  must 
be  considered  negative  when  estimated  in  a  contrary  direction. 

A  B  C 


Let  a  distance,  AB,  estimated  towards  the  right  from  the  point  A, 
be  represented  by  -f-  m ;  and  let  a  distance,  BC,  estimated  towards 
the  right  from  the  point  B,  be  represented  by  -f  ?i. 

Then,  AC=:AB -f  BC  =  TO  +  «. 

Now,  suppose  the  point  C  be  made  to  fall  between  A  and  B. 

AC  B 


Then,  AC  =  AB  —  BC  =  m  —  n . 

And  we  see  that  the  expression  for  BC  has  become  negative,  and 
that  this  change  of  sign  corresponds  to  a  change  in  the  direction  of 
BC.  It  was  first  estimated  towards  the  right  from  B;  it  is  now  esti- 
mated towards  the  left  from  B. 


NEGATIVE    QUANTITIES.  10?) 

Now,  suppose  BC  be  made  greater  than  AB,  tlieii  the  point  C  will 
fall  on  the  left  of  A. 

C A B_ 

And  m  —  n,  the  expression  for  AC,  will  become  negative.  And  we 
see  that  the  expression  for  AC,  which  was  positive  by  hypothesis  when 
the  distance  was  estimated  on  the  right  of  A,  has  become  negative 
when  the  distance  is  estimated  in  a  contrary  direction. 

197.  The  question  now  arises,  as  to  what  interpretation  is  to  be  put 
upon  a  negative  result  when  it  appears  in  the  solution  of  a  problem. 

Let  it  be  required  to  find  a  number,  which,  when  added  to  6,  will 
give  a  result  equal  to  4.*  Then,  from  tlic  conditions,  we  have  the 
statement  cc  +  6  =  4.  Hence,  x  ^=  —  2.  Now,  the  problem  is 
purely  arithmetical,  and  in  arithmetic  all  numbers  are  regarded  as  posi- 
tive. Therefore,  the  solution  is  absurd  in  an  arithmetical  sense,  but  it 
is  true  in  an  algebraic  sense.  For  substituting  —  2  for  a;  in  the  equa- 
tion of  the  problem,  we  have  —  2  -f  6  ==  4,  a  true  equation.  The 
value  satisfies  the  equation  of  the  problem.  It  is,  therefore,  a  true 
answer  to  the  problem  in  an  algebraic  sense.  But  it  is  not  a  true 
answer  to  the  problem  as  stated,  for  we  were  required  to  find  a  number, 
which,  when  added  to  6,  would  give  a  sum  ecjual  to  4 ;  and  we  have 
really  found  a  number  which,  srdjfracfed  from  6,  gave  a  difference 
equal  to  4.  The  negative  solution  has  then  satisfied  the  equation  of 
the  problem,  but  has  failed  to  fulfil  the  conditions  as  enunciated.  The 
explanation  of  this  difficulty  is  simple,  when  we  return  to  the  interpre- 
tation put  upon  a  negative  quantity.  We  have  said  that  a  negative 
quantity  always  indicates  a  change  of  direction  or  character.  That 
change  must  be  marked  by  a  corresponding  change  of  condition  in  the 
statement  of  a  problem,  and  then  the  negative  solution  will  be  changed 
into  a  positive  one.  The  preceding  problem  must  be  changed  into  this : 
required  to  find  a  number,  which,  when  subtracted  from  6,  will  give  a 
difference  equal  to  4. 

Then,  G  —  ./■  =  4,  and  x  =  +  2. 

A  negative  solution  will  not,  then,  satisfy  a  problem  as  enunciated, 
but  will  be  the  true  answer  to  another  problem  in  which  there  has  been 
a  change  of  condition  corresponding  to  the  indicated  change  of  charac- 
ter. But  negative  solutions  can  be  best  explained  by  the  discussion  of 
the  "  Problem  of  the  Couriers." 
10 


110  NEGATIVE    QUANTITIES. 


PROBLEM  OF  THE  COURIERS. 

198.  Two  couriers  travel  on  the  same  road ;  the  forward  courier  at 
the  rate  of  h  uiilcs  per  hour,  and  the  rear  courier  at  the  rate  of  a  miles 
per  hour.  At  12  o'clock  they  are  separated  by  a  distance  of  m  miles. 
The  problem  is,  to  determine  how  much  time  will  elapse  before  they 
are  together,  and  also  the  point  of  meeting. 

Let  the  indefinite  line 

A  BO 


represent  the  road  on  which  they  are  travelling ;  A,  the  position  of  the 
rear  courier;  B,  of  the  forward  courier  at  1^  o'clock;  0,  the  unknown 
point  of  meeting,  and  x  the  required  time  of  meeting.  Then,  from  the 
conditions  of  the  problem,  ax  =  AO,  hx  =  BO,  and  AB  =r  m  ;  and 
since,  from  the  figure,  we  have,  AO  —  BO  =  AB  —  m,  we  get  the 

statement,  ax  —  hx^=m.      Hence,  ,r  = -.     The  distance  they 

are  apart  at  12  o'clock,  divided  by  the  difl"erence  of  their  rates  of  travel, 
will  give  the  number  of  hours  that  must  elapse,  after  12,  before  they 
are  together.  The  time,  multiplied  by  the  rate  of  travel,  a,  will  give 
the  distance,  AO,  from  A  to  the  unknown  point  of  meeting  0 ;  or  the 
time,  multiplied  by  the  rate  h,  will  give  the  distance,  BO.  If,  for  in- 
stance, they  are  separated  by  a  distsince  of  48  miles  at  12  o'clock,  and 
the  forward  courier  travels  at  the  rate  of  4  miles  per  hour,  and  the  rear 
courier  at  the  rate  of  6  miles   per  hour,  then,  m  =  48,  a  =  6,  and 

48 
h  =  4.     Hence,  x  =  - — ^—7  =  24  hours,      ax  :=  Q  .  24  =  144  miles 
b  —  4 

=  AO.      ia;  =  4  .  24  =  96  miles  =  BO.      Verification,  AB  +  BO 

==  48  +  96  =  144  =  AO. 

We  might  make  AO  the  unknown  quantity,  and  call  it  )/  ;  then,  BO 

=  y  —  m.    Then,  — ,  the  distance  the  rear  courier  will  have  to  travel, 

divided  by  his  rate  of  travel,  will  give  the  time  that  must  elapse  before 

he  will  overtake  the  forward  courier.     So, ' — - —  will  give  the  time  that 

the  rear  courier  will  be  travelling  before  he  is  overtaken.  And,  since 
these  are  but  difi"erent  expressions  for  the  same  time,  we  get  the  equa- 
tion ^- =  — .     Hence,  y  = j ;  and  supposing,  as  before,  that 

li  a  a  —  u 


X  E  O  A  T  I  V  E     QUANTITIES.  Ill 

G  .  4S 
a=  Q,  l>  =  A,  and  ??i  =  48,  we  have  _y  =  AO  =  -r- — 7  =  144  miles. 

Then,  —  =  ---  =  24  hours,  as  before. 
a  b 

199.  We  will  confine  our  discussion  mainly  to  the  expression,  .r  = 

-,  in  which  x  represents  the  unknown  time.     Since  the  value  of 

a  fraction  increases  as  its  denominator  decreases,  it  is  evident  tjiat,  by 
making  the  difference  between  a  and  b  very  small,  we  can  make  the 
time  veiy  great.  This  is  apparent,  too,  from  the  problem ;  for,  if  the 
rear  courier  travel  but  little  faster  than  the  forward,  he  will  gain  but 
little  upon  him.  Now,  if  the  denominator  be  made  the  smallest  possi- 
ble, the  fraction  will  be  the  greatest  possible.  Hence,  when  the  deno- 
minator is  zero,  which  results  from  making  b  =  a,  the  fraction  must 
have  the  greatest  possible  value;  or,  in  other  words,  an  infinite  value. 

Therefore,  x  =  —  =  infinity,  synabulizcd  by  the  character  00 .     It  will 

then  be  an  infinite  time  after  12  o'clock  before  the  couriers  will  be  to- 
gether; that  is,  they  will  never  be  together.  This  is  plain,  from  tho 
nature  of  the  problem ;  for,  if  the  couriers  are  separated  by  a  distance 
of  m  miles  at  12  o'clock,  and  travel  at  equal  rates  after  12,  they  must 
always  keep  at  the  same  distance  of  m  miles  apart. 

The  character  00  ,  then,  indicates  impossibility.  "Whenever  it  ap- 
pears, by  examining  the  equation  of  the  problem,  we  will  discover  an 
absurdity  arising  from  the  hypothesis  that  gave  the  infinite  result.  In 
this  ease,  infinity  was  the  result  of  the  hypothesis,  a  =  b.  "When 
a  =  h,  ax  is  equal  to  bx,  or  AO  =  BO :  a  part  equal  to  the  whole, 
which  is  absurd. 

TThe  character  :»  ,  then,  indicates  two  distinct  things,  viz. :  impossi- 
bility in  the  fulfilment  of  the  required  conditions,  and  absurdity  in  the 
conditions  themselves. 

200.  Now,  let  b  become  greater  than  a,  the  denominator  of  the  value 
of  X  becomes  negative,  and  the  whole  fraction  negative.  "What  docs 
this  solution  mean  ? 

The  sign  of  the  time  being  minus,  indicates  that  it  refers  to  past 
time,  because  in  the  statement  we  regarded  future  time  as  positive. 
It  is  plain,  also,  from  the  nature  of  the  problem,  that  the  time  of  the 
couriers'  being  together,  if  together  at  all,  must  be  past,  not  future. 
For  the  forward  courier  travelling  more  rapidly  than  the  rear  courier, 


112  N  i:  r.  A  T  1  V  F,     Q  U  A  N  T  I  T  I  E  S  . 

■will  never  be  ovcvtr.ken  by  him.  The  solution,  then,  fliils  to  apply  to 
the  problem  as  enunciated ;  but  it  is  a  true  solution  for  another  ques- 
tion, viz. :  how  long  before  12  o'clock  were  the  two  couriers  together '{ 
Lot  us  return   to  the  particular  problem  :  making  h  =  0  instead  of 

a  =  C.      Then,  m  =  48,  a  =  -1,  h  =  G,   and   x  =  ^^  =  —  24. 

a  —  h 

Now,  it  is  obvious,  that  the  forward  courier,  gaining  two  miles  per  hour 

on  the  rear  courier,  must  have  been  with   him  24  hours  before   12, 

otherwise  the  two  would  not  be  separated  by  48  miles  at  12. 

_,  .  .      .     ,.  cini 

Eeturning  to  the  equation  in  distance,  y  =  ■ -,  we  see  that  the 

distance,  AO,  also  becomes  negative  under  the  hypothesis  of  a<^h. 
This  solution  indicates  impossibility  for  a  point  0  in  advance  of  B,  be- 
cause, when  a  <^  h,  ax  is  also  <^  bx,  or  AO  <^  BO,  which  is  absurd. 
But  it  will  be  a  true  solution  for  a  point  0  in  rear  of  A. 

A  B 


0 

Because,  for  such  a  point,  the  relation  ax  <^  bx  will  be  true. 

Resuming  the  values  m  =  48,  a  =  4,  and  i  =  G,  we  get  y  =  A(J 
=r  —  96  miles ;  which  agrees  with  the  solution  above,  for  the  courier 
A,  24  hours  before  12  o'clock,  being  at  0,  must  be  96  miles  from  0 
at  12. 

When  the  unknown  quantity  was  time,  we  have  seen  that  the  nega- 
tive solution  indicated  impossibility  for  future  time,  but  gave  a  true 
answer  to  the  question  in  past  time.  When  the  unknown  quantity  was 
distance,  the  negative  solution  indicated  impossibility  for  forward  dis- 
tance, but  gave  a  true  answer  to  the  question  in  distance  estimated%iu 
the  opposite  direction.  In  general,  a  negative  solution  indicates  that 
the  problem,  as  enunciated,  cannot  be  solved,  but  gives  a  true  solution 
when  the  unknown  quantity  is  changed  in  character  or  direction. 

That  being  the  case,  a  negative  solution  may  always  be  changed  into 
a  positive  solution  by  changing  the  character  or  direction  of  the  quan- 
tity that  produced  the  negative  result.     The  negative  soKition,  ;<:  = 

-,  may  be  changed  into  a  positive  solution  in  three  ways.     First, 

by  changing  the  direction  of  the  time ;  that  is,  by  changing  -f  x  into 
—  X,  which  is  equivalent  to  asking,  how  long  before  12  o'clock  the  two 
couriers  were  together.     Then  the  equation,  ax  —  bx=  m,  becomts 


NEGATIVE     QUANTITIES.  ll^l 


hx  —  aa:  =  m,  or  a:;  = ,  a  positive  result,  because   h  is  greater 

b  —  a 

than  a.     Second,  by  changing  —  h  into  +  h,  which  is  equivalent  to 

turning  the  forward  courier  around  to  meet  the  rear  courier.     The 

equation  then  becomes  ./■  —  j.     The  diagram  corresponds  to  this ; 

for,  in  this  case,  the  point  O  Avill  be  between  A  and  B,  and  wo  will 
have  ax  +  6x  =  AO  +  BO  =  AB. 

A  0  B 


Third,  by  changing  +  a  into  —  a,  and  —  h  into  +  h,  which  is  equi- 
valent to  turning  both  couriers  around,  and  making  the  courier  B  go  in 
pursuit  of  the  courier  A,  the  equation  becomes  —  ax  +  hx  =  m,  or 

X  = ,  a  positive  result.     It  will  be  seen  that  the  results  in  the 

6  —  a 

first  and  third  cases  are  the  same,  as  tliey  obviously  (night  to  be. 

201.  Now,  let  m  =  0,  and  a  >  h.     Then,  x  =  — -  =  0.       The 

a  —  h 

quotient  of  zero  by  any  finite  quantity  Ls  plainly  zero;  for  tlie  numera- 
tor of  a  fraction  indicates  the  amount  to  be  divided,  and  the  division 
of  nothing  must  give  nothing.  Thus,  if  a  man  has  zero  dollars  to  di- 
vide among  three  persons,  the  share  of  each  will  plainly  be  nothing. 
In  the  present  instance,  the  soliition,  0,  shows  that  the  couriers  will  be 
together  at  a  zero  time  after  12;  in  other  words,  at  12  itself.  This 
ought  to  be  so;  for,  when  m  =  0,  they  are  together  at  12 ;  and,  since 
they  travel  at  unequal  rates,  ^hey  will  be  together  no  more. 

202.  Now,  let  m  =  0,  and  a  =  h.     Then,  x  =  — .    To  explain  this 

symbol,  it  is  necessary  to  notice  particularly  the  conditions.  The  con- 
dition, m  =  0,  places  the  two  couriers  together ;  the  condition,  a  —  h, 
makes  them  travel  at  equal  rates.  Being  together  at  12,  and  travelling 
at  equal  rates,  they  must  nlways  be  together.  They  will  then  be  to- 
gether at  1,  at  2,  3,  3i,  &c.     There  is,  then,  no  dctenniiuUe  time  at 

which  they  are  together.     Hence,  —  is  called  the  si/mhol  of  Itxhtermi- 

natlon.     It  does  not  indicate  that  no  solution  can  be  found,  but,  on  the 
contrary,  that  too  many  can  be  determined ;  and  the  indetermination 
consists  in  this,  that  any  one  of  the  infinite  solutions  will  answer  just 
10*  II 


114  NEGATIVE    QUANTITIES. 

as  well  as  any  other.  Suppose,  for  instance,  the  answer  to  the  ques- 
tion, who  discovered  America?  was,  an  inhabitant  of  Europe  some 
time  after  the  Christian  era.  The  answer  would  be  indeterminate,  be- 
cause equally  applicable  to  countless  millions. 

0^ 

U 

This  will  be  a  true  equation,  when  x  =  1,  10,  1000,  anything  what- 
ever. Hence,  we  say  that  x  is  indeterminate.  The  diagram,  also, 
shows  the  same  thing. 

A  0  0  0  0  0 


B 

For,  when  m  =  0,  the  points  A  and  B  become  the  same ;  and,  since 
a  =  h,  then,  AO  =  BO  in  all  positions  of  the  point  0. 

Thus,  it  is  shown  in  three  ways  that  j-  is  the  symbol  of  indetermi- 

nation.  The  symbol  also  arises  from  an  identical  equation ;  for,  when 
a  =  h,  and  vi  =  0,  we  have  ax  =  ax,  an  identical  equation,  in  which 
X  may  have  any  value  whatever. 

By  recumng  to  the  equation  in  y,  we  will  observe  that  the  distance, 
from  the  common  point  of  starting  to  the  unknown  point  of  meeting, 
also  becomes  indeterminate  when  in  =  0  and  a  =  h.  This,  obviously, 
ought  to  be  so. 

203.  There  is  one  remarkable  exception  to  the  foregoing  statement, 
that  —  is  the  symbol  of  indetermination.     It  is  in  the  case  of  vanishing 

fractions.  A  vanishing  fraction  is  one,  which  becomes  — ,  in  conse- 
quence of  the  existence  of  a  common  factor  to  the  numerator  and  deno- 
minator, which  has  become  zero  by  a  particular  hypothesis  made  upon  it. 

204.  Take  the  expression  -— =  =  — ,  when  m  =  n,  but,  bv  de- 

m^  — n'^        0  '        '     J 

composing  the  denominator  into  its  factors,  we  have . 

(?n  —  n)  (ni  +  n) 

We  see  that  —  is  caused  by  the  common  factor,  m  —  n,  becoming  0 
by  the  hypothesis  m  =  n.     Divide  out  this  factor,  and  we  have  left 

=  r- ,  when  m  =  n. 

m  -{■  n       2n 


NEGATIVE    QUANTITIES.  115 

,      „       .        ???^  —  n^       (m  +  n)  (m  —  n)        0       , 

Ajrain,  take  the  traction, =  ^^ ■ — -  =  — -,  wnou 

'^  m  —  n  m  —  n  U 

m  =  n.     But,  divide  out  m — n,  and  we  have  '■ — - —  =  2n,  when 

„     (m  —  7if       (m  —  n)  (m  —  n)        0 

m  =  n.     So,  -^ = --^ =  -rr,  when  m  =  n.  But 

VI  —  71  m  — 11  0 

,  m  —  '^  r.  -I  c^ 

by  division,  the  fraction  becomes  — - —  =  U,   when    m  =  n.       ho. 

m   /M  mt  «.  Q 

=  — ,  when  m  =  n.      But,  by  divi- 


{m  —  nf       (m  —  «)  {m  —  n)        0 

111 

sum,  we  cret =  —  =  x,  when  m  =  n. 

We  conclude,  that  a  vanishing  fraction  has  one  of  three  trae  values; 
that  it  may  be  either  finite,  zero,  or  infinity.     How,  then,  are  wc  to 

decide  whether  the  symbol  —  indicates  indetcrmination,  or  a  vanishing 
fraction  ? 

If  the  particular  hypothesis  which  gives  the  symbol,  does  not  make 

the  given  equation  an  identical  equation,  we  may  be  certain  that  — 

points  out  a  vanishing  fraction. 

We  will  resume  the  subject  of  vanishing  fractions  more  at  lengt^i 
hereafter, 

205.  The  foregoing  symbols  are  of  the  highest  importance,  and  ought 
to  be  remembered. 

Arrangement  in  tabular  furm  will  assist  the  meiuory. 

m 

that  is,  a  finite  quantity,  divided  by  zero,  equal  to  infinity.  The  sym- 
bol, oo,  indicates  impossibility  in  the  fulfilment,  and  absurdity  in  the 
conditions  of  the  problem. 

that  is,  zero,  divided  by  any  finite  quantity,  equal  to  zero.  This  is  a 
true  solution,  unless  it  conflict  with  the  condition  of  the  problem. 

a;  =  —  A. 

The  negative  solution  indicates  that  the  problem,  as  enunciated,  can- 


116  GENERAL    PROBLEMS. 

not  be  solveil,  but  the  negative  result  will  be  a  true  answer  when  tlie 
unknown  quantity  is  changed  in  character  or  direction. 

0 

the  symbol  of  indetermination,  when  there  is  no  common  factor  in 
the  numerator  and  denominator.  By  indetermination  is  meant,  that 
there  is  an  infinite  system  of  values  that  will  satisfy  the  conditions  of 
the  problem. 

GENERAL  PROBLEMS. 

206.  1.  A  person  employed  a  workman  for  m  days,  upon  condition 
that  the  workman  should  receive  n  dollars  every  day  that  he  worked, 
and  should  forfeit  p  dollars  every  day  that  he  vras  idle ;  at  the  end  of 
the  time  he  received  c  dollars.  How  many  days  did  lie  work,  and  liuw 
many  was  he  idle  ? 

Let  X  =  working  days,  then  m  —  x  =  idle  diiys;  and  by  the  con- 
ditions, wc  get  )tx  —  (in  —  .t)  jj  =  c. 

Hence,  x  —  -,  on  —  x  =  idle  days  =  m  — — . 

n  +  J)  n  +2)         n  +  p 

When  will  the  last  expression  be  zero,  and  when  negative,  and  v/^hat 
do  these  solutions  indicate  ? 

The  negative  solution  needs  some  explanation  :  o,  the  entire  amount 
received,  cannot  exceed  nm,  what  was  paid  for  working  ni  days,  unless 
the  workman  had  not  only  no  forfeiture  to  pay,  but  also  worked  more 
than  the  time  for  which  he  was  employed.  The  idle  days  have  then 
changed  their  character,  and  become  working  days  beyond  the  period 
for  which  the  man's  services  were  engaged. 

2.  The  same  problem  as  before,  except  the  workman,  at  the  end  of 
the  time,  was  in  debt  c  dollars. 

,  77?;)  —  c        ,  nm  -f-  <• 

Alls.  X  —  —i- ,  and  m  —  x  = . 

n  +  m  n  -f  p 

When  will  the  first  solution  become  zero,  and  when  negative  ? 

3.  The  same  problem,  except  that  the  workman  received  nothing. 

mp  ,  nm 

Ans.  X  =  — - — ,  and  m  —  ./•  =  . 

11  +  2^  '^  +  1^ 

The  above  results  are  formulas,  and  may  be  made  applicable  to  par- 
ticular problems  of  the  same  kind. 

4.  In  a  certain  college,  the  maximum  mark  for  a  recitation  is  100. 


GENERAL    PROBLEMS.  117 

A  Freshman  recited  5  times  in  mathematics;  two  days  he  got  95 
for  each  recitation,  and  his  average  for  the  5  days  was  86.  "What  did 
lie  average  each  of  the  other  three  days  ?  Ans.  80. 

5.  Divide  the  number  a  into  two  such  parts  that  the  product  of  the 
first  by  m  shall  be  equal  to  the  quotient  arising  from  dividing  the 
second  by  n. 

a  -         071711 

Ans.   = ,  and  ■= . 

1  +  117)1  1  +  nm 

What  effect  has  increasing  n  upon  both  results?  Under  what  form 
must  the  second  expression  be  placed  to  show  that  it  is  directly  propor- 
tional to  n  and  mP     When  will  the  two  expressions  be  equal? 

6.  A  father  is  32  years  old,  and  his  son  8  years.  How  long  will  it 
be  until  the  age  of  the  father  is  just  double  that  of  the  son  ?  • 

Ans.  16  years. 

7.  A  Either  is  a  years,  and  his  son  h  years  old.     How  long  until  the 

age  of  the  father  will  be  c  times  as  great  as  that  of  the  son  ? 

.  a  —  ch 

Ans.   :r  = —. 

c  —  1 

What  do  those  values  become  when  c  =  1  ?  What,  when  ch  =  a? 
What,  when  ch^  a'i  What  do  these  different  solutions  indicate ? 
When  c  —  \,  x=  cc,  the  symbol  of  absurdity,  as  it  ought  to  be. 
The  hypothesis  makes  the  father  ami  son  equal  in  age  after  the  lapse 
(if  h  years.  When  ch  =  a,  x  =  0.  A  true  solution,  since  the  father 
is  now  c  times  as  old  as  his  son.  When  cb'^  a,  x  is  negative.  Fu- 
ture time  being  positive,  past  time  is  negative.     The  solution,  then, 

indicates  that r-  years  ago,  the  father  was  c  times  as  old  as  his 

son. 

8.  A  father  is  64,  and  his  son  16.  How  long  will  it  be  until  the 
age  of  the  father  is  9  times  as  great  as  that  of  the  son. 

Ans.  X  =  —  10. 
The  solution  indicates  that  10  j-ears  ago  the  age  of  the  father  was  9 
times  as  great  as  that  of  the  sun.     The  father  was  then  54,  and  the 
son  6  years  old. 

9.  Bronze  cannon  (commonly,  but  improperly,  called  brass  cannon) 
are  composed  of  90  parts  of  copper,  and  10  parts  of  tin :  8j%  lbs.  of 
copper  lose  1  lb.  when  immersed  in  water,  and  7j%  lbs.  of  tin  lose  1  lb. 
when  immersed  in  water.  The  Ordnance  Board  suspecting  that  some 
bronze  cannon  did  not  contain  copper  enough,  immersed  one  of  them, 


118  GENERAL    PROBLEMS. 

weighing  000  lbs.,  aud  found  its  loss  of  weight 
What  was  its  composition  ? 

Ajis.   so  parts  of  copper,  and  10  parts  of  tin. 
Verify  the  results.     The  copper  lost  89^;J  lbs.,  aud  the  tin  13 y  lbs. 
The  sum  of  which  is  1033|0  9  lbs. 

10.  A  fox,  pursued  by  a  greyhound,  is  125  of  his  own  leaps  ahead 
of  the  greyhound,  and  makes  6  leaps  to  the  greyhound's  5,  but  2  of 
the  greyhound's  leaps  are  equal  to  3  leaps  of  the  fox.  How  many  leaps 
will  the  fox  make  before  he  is  caught  by  the  greyhound  ? 

Let  X  =  number  of  leaps  made  by  the  fox;  then,  since  the  grey- 
hound makes  but  5  leaps  while  the  fox  makes  6,  he  will  make  |  cf  a 
leap  to  one  leap  of  the  fox.  Therefore,  he  will  make  |.x  leaps'  while 
the  fox  makes  x  leaps.  But  each  of  the  greyhound's  leaps  are  equal 
to  I  of  a  leap  of  the  fox.  Hence,  the  |x  leaps  of  the  greyhound  are 
equivalent  to  |  .  ^x  leaps  of  the  fox ',  and  since  the  greyhound  has  not 
only  to  run  over  the  ground  passed  by  the  fox  in  making  x  leaps,  .but  ,. 
also  that  passed  in  making  125  leaps,  we  have  the  equation  of  tlie  pro- ,,' 
blem.     I  .  |x  =  :k  +  125.  Ans.  x  =  500  leaps,      ;:, 

Verification.     While  the  fox  made  500  leaps,  the  greyhound  made/. 
4161  leaps ;  and  these  were  equal  to  |  .  4161,  or  625  leaps  of  thq,  fox, 
the  entire  number  of  leaps  made  by  the  fox. 


Bcmarks. 

This  problem  shows  the  importance  of  making  the  two  members  ex- 
press the  same  thing  by  referring  them  to  the  same  unit.  The  leaps 
of  the  greyhound  have  been  expressed  in  terms  of  those  of  the  fox. 
We  might  have  made  the  unknown  quantity  represent  the  number  of 
leaps  made  by  the  greyhound,  and,  in  that  case,  the  leaps  of  the  fox 
must  have  been  expressed  in  terms  of  those  of  the  greyhound. 

11.  A  fox,  pursued  by  a  greyhound,  has  a  start  of  a  leaps,  aud 
makes  h  leaps  while  the  greyhound  makes  c  leaps ;  but  d  leaps  of  the 
greyhound  are  equal  to  e  leaps  of  the  fox.  How  many  leaps  will  tlie 
fox  make  before  he  is  overtaken  by  the  greyhound  ? 

,  adb 

^''s.   X  =  ^^_^^. 

When  will  x  =  oP  When  will  it  be  equal  to  infinity?  When  will  it 
be  negative  ?     How  are  these  solutions  explained  ? 

X  will  be  zero,  when  a  is  zero,     x  will  be  infinite  when  ec  =  Jh ; 


GENERAL    PROBLEMS.  119 

that  is,  when  the  number  of  leaps  made  by  the  greyhound,  multiplied 
by  the  value  of  each  leap,  is  equal  to  the  number  of  leaps  of  the  fox, 
multiplied  by  the  value  of  each  of  his  leaps.  In  that  case,  the  hound 
•will  evidently  never  overtake  the  fox  ;  and  the  solution  indicates  impos- 
sibility or  absurdity,  x  Tvill  be  negative  when  db  is  greater  than  ec ; 
then  the  fox  is  running  faster  than  the  hound,  and  the  distance  re- 
presented by  the  x  leaps  must  be  estimated  in  a  contrary  direction. 

P'  11  F  P 


Let  II  be  the  position  of  the  hound ;  F,  that  of  the  fox ;  and  P  the 
point  where  the  fox  is  overtaken  by  the  hound.  Then,  when  x  is  ne- 
gative, we  understand  either,  that  the  fox  pursued  the  hound  and 
caught  up  to  him  at  some  point,  V  on  the  left^  or  that,  at  some  time 
previous  to  the  fox  being  at  F,  and  the  hound  at  H,  they  were  both 
together  at  P',  and  the  fox  ruuiiiug  faster  than  his  pursuer  gained  upon 
him  the  distance  HF  — -  a. 

12.  A  man,  desirous  of  giving  4  cents  apiece  to  some  beggars,  found 
that  he  had  not  money  enough  by  5  cents ;  lie  therefore  gave  them  2 
cents  apiece,  and  had  15  cents  left.  IIow  many  beggars  were  there, 
and  how  much  money  had  he  ?  Ans.   10  beggars  :  35  cents. 

13.  A  man,  desirous  of  giving  n  cents  apiece  to  some  beggars,  found 
that  he  had  not  money  enough  by  h  cents ;  he  therefore  gave  them  c 
cents  apiece,  and  had  d  cents  left.  IIow  many  beggars  were  there, 
and  how  much  money  had  the  man  ? 

d  +  h  ^  ^nd  -]-  rh 

Ans.  bcirirars,  and  cents. 

a  —  c      " ""  a  —  f 

What  values  must  a,  I,  c,  and  d  have  to  make  this  problem  the 
same  as  the  last  ? 

When  c  ]]>  a,  the  solution  is  negative;  and  negative  solutions  indi- 
cate a  change  of  direction  and  character.  The  beggars  become  givers  : 
the  amount  given  becomes  the  amount  received :  the  deficiency  be- 
comes a  surplus,  and  the  surplus  a  deficiency.  When  c  —  a,  both  so- 
lutions become  infinite,  and  the  symbol,  cc,  here  plainly  indicates 
absurdity. 

When  d  and  L  are  both  zero,  the  two  solutions  will  be  both  0,  or 

§-•     Why? 

14.  A  man,  desirous  of  giving  2  cents  apiece  to  some  beggars,  found 
that  he  had  not  money  enough  by  15  cents ;  he  therefore  gave  them 


120  O  K  N  E  R  A  I-     1'  R  C)  B  L  EMS. 

4  cents  jipicce,  and  liad  5  cents  left.     How  many  beggars  were  there, 
and  how  much  money  had  tlie  man  ? 

Ann.  —  10  beggars,  and  — •  S5  cents. 
How  is  the  solution  explained?     There  were  10  givers,  who  each 
gave  the  man  2  cents,  and  15  cents  over;  or  each  4  cents  apiece,  lack- 
ing 5  cents  in  all. 

15.  A  piratical  vessel  sails  at  the  rate  of  r  miles  per  hour  for  a 
hours,  when  she  sustains  some  injury,  and  can  only  sail  /  miles  per 
hour.  At  the  moment  in  which  the  piratical  vessel  is  disabled,  a 
sloop-of-war  starts  in  pursuit,  and  sa*ls  at  the  rate  of  r  miles  per  hour, 
from  the  point  where  the  pirate  first  started.  How  long  before  the 
sloop  will  overtake  the  pirate  ? 

Ans.  x  = >  hours. 

J.  —  / 

When  will  this  solution  be  zero?  When  negative?  When  infinite? 
Under  what  form  must  the  fraction  be  placed  to  show  that  x  is  recipro- 
cally proportional  to  r  ? 

Make  the  general  solution  applicable  to  a  particular  example. 

16.  Divide  a  number,  a,  into  two  parts,  which  shall  be  to  each  other 

as  m  is  to  n. 

ma  ^       va 

A)is.   ,  and 


4-  11  111  +  n 

17.  Divide  a  number,  a,  into  three  parts ;  such,  that  the  first  shall  be 

to  the  second  as  7i  to  m,  and  the  second  to  the  third  as  q  to  j5. 

aiiq  amq  amp 

Ans.    , ) . 

mp  +  mq  -f  nq      mp  -f  mq  +  nq      mp  +  mq  +  nq 

What  supposition  will  make  all  the  parts  zero  ?  What  one  of  them  ? 
What  two  of  them  ? 

18.  Divide  a  number,  a,  into  four  parts;  such,  that  the  first  shall 
be  to  the  second  as  n  to  m,  the  second  to  the  third  as  q  to  p,  and  the 
third  to  the  fourth  as  r  to  s. 

anqr  amqr 

nqr  +  mqr  +  mpr  +  mp)s      nqr  -f  mqr  +  mpr  -f-  mps^ 

ampr  omps 


i\qr  +  mqr  +  mpr  -f  mps      nqr  +  mqr  +  mpr  +  mp 

19.  Milk  sells  in  the  City  of  New  York  at  4  cents  per  quart.  A 
milkman  mixed  some  water  with  50  gallons  of  milk,  and  sold  the  mix- 
ture at  3  cents  per  quart  without  sustaining  any  loss  by  the  sale.  How 
much  water  did  he  put  in  the  milk?  An^.   GG§  quarts. 


GENERAL    TROBLEMS.  121 

20.  Milk  sells  ia  Boston  at  a  cents  per  quart.     A  milkman  mixed  a 

certain  quantity  of  water  with  b  quarts  of  milk,  and  sold  the  whole  at 

(•  cents  per  quart  without  losing  anything  hy  the  sale.     How  much 

water  was  added  to  the  milk  ? 

ah  —  he 
Ans.  j:  = —  quarts. 

The  value  of  x  is  zero  when  c  =  a.  In  that  case,  evidently,  no 
water  is  added.  The  value  is  infinite  when  c  =  o.  Then  the  milkman 
gives  away,  gratuitously,  an  infinite  quantity  of  water.  "When  c  ^  h, 
X  is  negative.  '  Then  we  understai^l  that  a  certain  quantity  of  water  is 
separated  from,  not  added  to  the  milk,  and  that  the  price  of  the  milk  is 
c  cents  per  quart,  and  of  the  mixture  a  cents  per  quart. 

21.  How  often  are  the  hour  and  minute  hands  of  a  clock  together? 

An>>.  J^very  Qbj^j  minutes. 

22.  How  often  are  the  minute  and  second  hands  of  a  clock  to- 
gether? Ans.  Every  13'g  minutes. 

23.  How  often  are  the  hour,  minute,  and  second  hands  of  a  clock 
together  r  Ans.  Every  720  minutes. 

The  last  problem  is  solved  by  means  of  the  least  common  multiple. 

24.  There  is  an  island  60  miles  in  circumference.  Three  pereons 
start  from  the  same  point  to  travel  around  it,  travelling  at  the  respect- 
ive rates  of  4,  6,  and  1(5  miles  per  hour.  How  often  will  all  three  be 
together?  Ans.  Every  30  hours. 

This  problem  is  solved  by  means  of  the  least  common  multiple. 

25.  There  is  an  island  a  miles  in  circumference,  around  which  three 
persons  start  to  travel,  at  the  rates  of  h,  c,  and  d  miles  per  hour. 
When  will  they  all  be  together  again  ? 

Alls.  In  a  hours,  divided  by  the  greatest  common  divisor  of  c  —  h, 
and  d  —  c. 

26.  Same  problem  as  24,  except  that  there  are  4  persons  travelling, 
at  the  respective  rates  of  3,  6,  12,  and  27  miles  per  hour. 

Ans.  Every  20  hours. 

27.  A  farmer  purchases  a  tract  of  land  for  $500,  on  a  credit  of  10 
months,  or  S480  cash.  What  is  the  rate  of  interest  that  makes  these 
sums  equivalent.  Ans.  5  per  cent. 

Let  .T  =  interest  upon  $100  fur  one  month.     The  statement  will  be 
100  +  lO.r  :  100  :  :  500  :  480. 
11 


122  GENERAL     PROBLEMS. 

28.  A  farmer  buys  a  tract  of  laud  for  a  dollars,  payable  iu  h  moutlis, 

or  for  c  dollars  casli.     What  is  the  rate  of  iutercst  i* 

100  (a  —  c) 

Ans.   X  = ^-^ -. 

c/j 

What  suppositiou  will  make  this  value  zero?  What  two  supposi- 
tious will  make  it  infiuite  ?  What  supposition  will  uiake  it  negative  ? 
and  how  is  the  negative  solution  explained  ? 

29.  Two  numbers  are  to  each  other  as  8  to  3 ;  but  if  8  be  added  to 
both  numbers,  the  first  will  only  be  double  the  second.  What  are  the 
numbers?  Ans.  32  and  12. 

30.  Two  numbers  are  to  each  other  as  a  to  h  ;  but  if  c  be  added  to 
both  of  them,  the  first  will  only  be  d  times  as  great  as  the  second. 

What  are  the  numbers  ? 

ac(d—l)  fjcQj  —  1) 

Ans.  First,  ^ 7-,—  :  second  — ^^ r-~. 

a  —  bd  a  —  Ixl 

What  do  these  solutions  become  when  cZ  =  1  ?  What,  when 
hd  =  a?     What,  when  bd'^  al     What,  when  d  =  1,  and  hd  =  a  ? 

The  first  hypothesis  gives  a  true  solution.  The  second  gives  an  ab- 
surd solution,  as  it  ought,  since  hd  can  only  equal  a  when  the  first 
number,  after  the  addition  of  c  to  both  numbers,  exceeds  the  second 
proportionally  as  much  as  before.  But  this  is  impossible,  since  the 
smaller  increases  most  rapidly.     See  Article  96. 

The  hypothesis,  hd,  ^  (/,  gives  a  solution  impossible  for  arithmetical 
quantities,  but  possible  for  algebraic.  We  must  either  suppose  that 
two  numbers  are  to  be  found,  which  result  from  the  subtraction  from  a 
number  not  expressed,  or  we  must  change  the  character  of  the  problem, 
and  make  c  subtractive.  When  hd  =  a,  and  r?  :=  1,  the  two  numbers 
are  represented  by  g,  the  symbol  of  indetermination.  An  infinite,  or 
indeterminate,  number  of  quantities  will  satisfy  the  conditions  of  the 
problem. 

The  following  problems  will  illustrate  the  foregoing  cases  : 

31.  Two  numbers  are  to  each  other  as  4  to  3 ;  but  when  5  is  added 
to  both,  the  numbers  are  equal.     That  is,  tZ  =  1 . 

Ans.  Both  numbers  zero. 

32.  The  same  as  last,  except  that,  after  the  addition,  the  first  will  be 
I  greater  than  the  second.     That  is,  hd  =  a. 

I  Ans.   Both  infinite. 


GENERAL    PROBLEMS.  123 

33.  The  same  as  31,  except  that,  after  the  additions,  the  first  will  be 
twice  as  great  as  the  second.     That  is.  Id  ^  a. 

Ans.  —  10,  and  —  11. 

Change  the  character  of  the  problem,  and  take  5  from  both  numbers, 
and  the  solutions  will  be  +  10  and  +  7. 

34.  The  same  as  31,  except  that  the  numbers  were  equal  before  the 
addition  (5f  5  to  both. 

Ans.  Both  indeterminate,  — .       An}-  numbers  will  answer. 

35.  A  fiither  divided  his  estate,  worth  $1200,  among  his  three  sons, 
so  that  the  share  of  the  first  should  be  to  that  of  the  second  as  6 
to  4 ;  and  so  that  the  share  of  the  third  should  be  the  greatest  common 
divisor  of  the  shares  of  the  first  and  second.  What  will  be  the  share 
of  each  ?  Am.  6600,  $400,  and  $200. 

30.  A  father  divides  his  estate  among  his  three  sons,  so  that  the 

share  of  the  first  shall  be  to  that  of  the  second  as  —     is  to  '-—  :    and  so 

q  s 

that  the  share  of  the  third  shall  be  the  greatest  common  divisor  of  tht- 

other  two.     The  father's  estate  is  worth  a  dollars :  what  is  the  share 

of  each  son  ? 

,            miq  ,  .   ,  a 

second,  — ;  tmrd. 


VIS  +  H J  +  1 '  '  ms  -\-  nq  -\-  \  ms  +  nq  +  1 

When  will  the  shares  of  the  first  and  second  be  equal  ?     What  will 
the  share  of  the  third  be  then  ?     What  will  be  the  efiect  upon   the 


37.  The  difference  of  two  numbers  is  a,  and  the  difference  of  their 
squares  is  zero.     What  are  the  numbers  ? 

38.  The  sum  of  two    numbers  is  2a,  and  the  difference  of  their 

squares  equal  to  c.     What  are  the  numbers  ? 

4a^  +  c         ,  4«-  —  r 
Ans.    — ,  and 


4«     '  4« 

What    do    these    solutions    become    when    c  ^  0  ?      What,    when 
c  =  4a'?     What,  when  r  ^  4a-? 


124 


G  E  N  E  R  A  I.     1'  R  ()  B  L  E  M  S  . 


n  such  a  manner  that  the  track  of  one  cuts  the 
parallel  of  latitude  through  the  Island  of  St. 
Helena  40  miles  on  the  west  of  that  island, 
and  cuts  the  meridian  through  the  island  40 
miles  north  of  it;  and  that  the  track  of  the 
other  cuts  the  parallel  of  latitude  80  miles  on 
the  east,  and  the  meridian  80  miles  on  the 
north  ?      Where  will   the   two  tracks  intersect 


each  other  ? 


Ans.  20  miles  east,  and  GO  miles  north  of  the  island. 
Call  DO,  X  ;   then,  PJ)  =  ./■  +  40  ;    and  PD  =  80  —  a:      Hence, 
40  +  .r  =  80  - 


then,  ^  =  20  ;  and  PD  =  60. 


40.  A  Yankee  mixes  a  certain  number  of  wooden  nutmegs,  which 
cost  him  I  cent  apiece,  with  a  quantity  of  real  nutmegs,  worth  4  cents 
apiece,  and  sells  the  Avhole  assortment  for  $44;  and  gains  $3-75  by  the 
fraud.     How  many  wooden  nutmecs  were  there  ? 

Ans.  100. 

41.  A  Yankee  mixed  a  certain  quantity  of  wooden  nutmegs,  which 

1th 
cost  him  — —  part  of  a  cent  apiece,  with  real  nutmegs,  worth  c  cents 

apiece,  and  sold  the  whole  for  a  dollars.     He  gained  d  cents  by  the 
fraud.     How  many  wooden  nutmegs  were  there  ? 

What  does  this  become  when  f ?  =  0  ?  What,  when  h  =  {)'{  What, 
when    he  =  11     What,  when  he  <^1. 

42.  The  sum  of  three  numbers  is  200 ;  the  first  is  to  the  second  as 
5  to  4,  and  the  thii'd  is  the  greatest  common  divisor  of  the  first  two. 
What  are  the  numbers  ?  Ans.   100,  80,  and  20. 

43.  The  sum  of  three  numbers  is  a  ;  the  first  is  to  the  second  as  h 
to  0,  and  the  third  is  the  least  common  multiple  of  the  first  two.  What 
are  the  numbers  ? 

ah  ac  nhc 


Am 


h  -{■  c  +  he      6  -f  (•  +  he      6  +  c  +  he 
What  single  hypothesis  will  make  them  all  equal  ?     What  will  these 
values  become  when  a  —  115,  h  =  25,  and  c  =  I'o. 

41.  A  northern    railroad    company  is    assessed    $120,000    damages 
for  contusions  and  broken  limbs,  caused  by  a  collision  of  cars.     They 


GENERAL    PROBLEMS.  125 

pay  S5000  for  each  coutusiou,  aud  S6000  for  each  broken  limb ;  and 
the  entire  amount  paid  for  bruises  and  fractures  is  the  same.  How 
many  persons  received  contusions,  and  how  many  had  their  limbs 
broken  ?  Atis.  12  of  the  former,  and  10  of  the  latter. 

45.  Same  problem  as  the  last,  except  representing  the  assessment  by 

a,  the  price  of  each  contusion   by  c,  and  that  of  each  broken   limb 

hjh. 

a  (I 

A)is.    —,  and  — . 

'Sc  -u 

What  do  these  values  become  when  c  =  0  y  What,  when  h  =  0  ? 
What,  when  a  =  0  ?     What,  when  h  ■=^  c? 

4G.  The  reservoir  at  Lexington  contains  48,000  gallons  of  water, 
and  supplies  the  town  and  Virginia  Military  Institute.  If  all  the  con- 
ducting pipes  were  closed,  the  reservoir  would  supply  the  town  and 
the  Institute  for  57|  hours,  aud  the  town  alone  for  96  hours.  How 
many  gallons  docs  the  Institute  use  per  hour. 

Ans.  333  J  gallons. 

47.  Two  pipes  will  exhaust  a  cistern  containing  a  quantity  of  water, 
represented  by  q,  in  a  hours,  and  the  first  will,  alone,  exhaust  it  in  h 
hours.  How  long  will  it  take  the  second  pipe  to  empty  it,  and  how 
much  docs  it  exhaust  per  hour. 

Ans. hours,  and ^    —  gallijus  per  hour. 

6  —  a  ab 

What  do  these  solutions  become  when  d  =  f>,  and  a~^  h? 

The  negative  solution  can  be  illustrated  by  an  example. 

48.  Two  pipes,  a  and  h,  will  exhaust  a  cistern  in  4  hours;  and  the 
pipe  a  can,  alone,  exhaust  it  in  2  hours.  In  what  time  can  the  pipe  h 
empty  the  cistern  ?  Ans.  In  —  2  hours. 

The  solution  being  negative,  may  refer  to  past  time,  and  indicate 
that,  during  the  two  hours  before  the  pipe  a  began  to  play,  the  pipe  // 
had  exhausted  the  cistern.  The  two  hours  play  of  the  pipe  h,  added 
to  the  two  of  the  pipe  a,  give  the  4  hours  of  the  problem.  Or,  we 
may  suppose  that  the  character  of  the  pipe  h  has  been  changed,  and 
that  it  has  been  a  supplying  pipe  for  two  hours,  and,  consequently,  the 
pipe  a  has  been  twice  the  time  in  exhausting  the  cistern. 

49.  A  man  asks  $120  cash  for  his  horse,  or  §126-30  on  a  credit  of 
9  months.  Supposing  these  valuations  to  be  equal,  what  is  the  rate  of 
interest  ?  Ans.  7  per  cent. 

11* 


126  GENERAL    PROBLEMS. 

50.  A  man  values  liis  horse  at  c  dollars  cash,  or  h  dollars  on  a  credit 
of  a  months.     What  is  the  rate  of  interest  ? 

1200  (i  —  c) 

Ans.  X  = ^: -. 

ac 

"What  does  x  become  when  h  =  c?     What,  when  a'^  h?     What, 

when  or,  or  c  =  0  ? 

51.  There  is  an  island  32  miles  in  circumference  :  two  persons  start 
to  travel  around  it,  the  first  at  the  rate  of  11  miles  per  day,  and  the 
second  at  the  rate  of  3  miles  per  day.  When  will  the  distance  be- 
tween them  be  equal,  estimated  in  either  direction  around  the  island  ? 

Ans.  At  the  end  of  the  second  day. 

52.  Two  persons  start  to  travel  around  an  island  a  miles  in  circum- 
ference, the  first  at  the  rate  of  h  miles  per  day,  and  the  second  at  the 
rate  of  c  miles  per  day.  When  will  the  distances  between  them  be 
equal,  estimated  in  both  directions  around  the  island  ? 

Ans.  X  =  - — ; 

2  (6  -  c) 

What  will  this  solution  become  if  the  travellers  move  at  equal  rates  ? 

What,  if  the  second  travels  the  fastest  ?     What,  if  the  second  stops  ? 

53.  Three  persons  start  to  travel  around  an  island  which  is  a  perfect 
circle,  120  miles  in  circumference.  The  first  travels'at  the  rate  of  4  miles 
per  day,  the  second  at  the  rate  of  8  miles  per  day,  and  the  third  at  the 
rate  of  6  miles  per  day.  How  many  days  will  elapse  until  the  lines 
joining  their  respective  positions,  and  the  centre  of  the  island,  will 
trisect  the  circumference,  under  the  supposition  that  the  second,  only, 
has  gone  past  the  starting  point  on  a  second  tour  of  the  island  ? 

Ajis.  20  days. 

54.  Three  persons  travel  around  a  circular  island,  160  miles  in  cir- 
cumference. They  start  together,  and  travel  at  the  respective  rates  of 
7,  5,  and  2  miles  per  hour.  How  long  will  it  be  until  the  third  will 
be  between  the  first  and  second,  and  equally  distant  from  them,  under 
the  hypothesis  that  the  first  and  second,  only,  have  passed  the  starting 
point  ?  Avs.  40  hours. 

55.  Three  persons  travel  around  an  island  a  miles  in  circumference. 
They  travel  at  the  respective  rates  of  h,  c,  and  d  miles  per  hour.  How 
long  will  it  be  until  the  third  will  be  half  way  between  the  first  and 
second,  and  how  fiir  will  the  first  have  travelled  ? 

Ans.   Distance, r^-^— —  :  time, —^r-. —  • 

0  —  2c/  -f  c  h  —  2rt  -j-  c 


GENERAL    PROBLEMS.  127 

What  do  these  values  become  when  2d  =  h  -\-  c  ?  "What,  when 
2d  =  c?     How  do  you  explain  these  results ? 

5G.  Divide  the  number  24  into  two  such  parts  that  their  product 
be  the  greatest  possible. 

Let  X  express  the  excess  of  one  of  the  parts  over  the  half  of  12, 
then,  Q  -{-  X,  and  6  —  x  will  represent  the  two  parts.  And  (G  +  .r ) 
(6  —  a;),  36  —  x^  is  to  be  the  greatest  possible.  This  result  will  evi- 
dently be  the  greatest  possible  when  a-  =  0,  or  when  the  parts  are 
equal. 

57.  Divide  the  quantity  a  into  two  such  parts  that  their  product 

shall  be  the  greatest  possible.     "What  are  the  parts  ? 

n  a 

Ans.    — ,  and   — . 

58.  Two  men  have  each  an  end  of  a  pole  0  feet  long  upon  their 
shoulders,  with  a  burden  suspended  from  it.  The  share  of  the  load  sus- 
tained by  each  man  is  inversely  proportional  to  the  distance  of  the  load 
from  his  shoulder.  The  proportion  of  the  weight  borne  by  one  man  is 
to  that  borne  by  the  other  as  3  to  2.  How  far  is  the  burden  from  the 
shoulder  of  the  first  man  ?  Ans.  T^  feet. 

59.  Same  problem  as  the  last,  except  the  representation  of  the  length 
of  the  pole  by  a,  and  the  proportion  by  h  and  c.  How  far  is  the  bur- 
den from  the  shoulder  of  each  man  ? 

^^      c  ,      (dj 

Ans. feet,  and feet. 

l>  +  c  b  -\-  c 

When  will  these  distances  be  equal  ?     "When  one  double  the  other  ? 

00.  A  general  wishing  to  range  his  men  in  a  solid  square,  found 
that  he  had  too  many  men  by  100.  He  increased  each  side  of  the 
square  by  one  man,  and  then  found  that  he  had  but  39  too  many 
How  many  men  had  he  ?  Ans.  1000. 

Let  J-  =  side  of  the  square. 

61.  A  general  ranged  his  army  in  a  solid  square,  and  found  that  he 
had  a  more  men  than  would  enter  the  square.  He  then  increased 
each  side  of  the  square  by  one  man,  and  found  that  he  had  h  more 
men  than  could  enter  the  square.  How  many  men  were  in  each  side 
of  the  square  ? 

A...  X  =  "-<*  + ^l 


128  GENERAL    PROBLEMS. 

What  does  this  value  become  when  6  +  1  =  a  ?  What,  when 
h  =  —  1  ?  How  do  you  expkiin  these  results  ?  The  last  result  will 
confirm  a  truth  hereafter  to  be  demonstrated. 

62.  Divide  the  number  20  into  three  such  parts  that  the  half  of  the 
first,  the  one-third  of  the  second,  and  the  one  fifth  of  the  third  shall  be 
equal.     What  are  the  parts?  Ans.  4,  G,  and  10. 

For,  let  X  represent  the  equal  result  after  the  division  of  the  three 
parts  by  2,  3,  and  5;  the.  parts  themselves  will  plainly  be  represented 
by  '2x,  3.r,  and  5x.  Hence,  2a;  +  3.7;  +  bx  =  20,  or  j:  =  2.  Then, 
2a;  =  4,  Sa-  =  6,  and  5x  =  10. 

63.  Divide  the  number  a  into  three  such  parts  that  the  first  divided 
by  h,  the  second  by  c,  and  the  third  by  cl,  shall  all  be  equal.  What 
are  the  parts  ? 

.  ah  ac  ,  ad 

^--  M^TTrf-   hTTTd' ^'^'^  h+v^r 

When  will  the  three  parts  be  equal  ?  What  will  be  the  eifect  of 
making  &  =  0  ?     How  do  you  explain  the  result? 

64.  Two  laborers  are  engaged  to  dig  a  ditch ;  the  first  can  dig  it  in 
5  days,  the  second  in  3i  days.  How  long  will  it  take  the  two,  working 
together,  to  dig  the  ditch  ?  Ans.  2  days. 

Verify  the  result. 

65.  Two  men  are  employed  to  dig  a  ditch ;  the  first,  alone,  can  dig 

it  in  a  days,  and  the  second,  alone,  can  dig  it  in  h  days.     How  long 

will  it  take  the  two,  laboring  together,  to  perform  the  work  ? 

,  <xh 

Aim.   X  = . 

b  +  a 

When  will  the  time  be  one  half  of  that  required  by  the  first  to  dig 
it  ?  What  change  will  this  solution  undergo  when  the  second  man  fills 
up  instead  of  digs  out  ?  What  will  be  the  expression  for  the  time  in 
that  case,  when  he  fills  up  as  fast  as  the  first  digs  out  ? 

This  problem  shows  clearly  that  a  change  in  a  c(jndition  will  be 
accompanied  by  a  change  of  sign, 

66.  A  debt  of  $150  was  paid  in  dollar  and  five-cent  pieces.  The 
dollar  and  five-cent  pieces  were  1100  in  number.  How  many  dollar, 
and  how  many  five-cent  pieces  were  used  ? 

Am.  100  dollar  pieces,  and  1000  five-cent  pieces. 

67.  A  debt  of  a  cents  was  paid  in  h  and  c  cent  pieces,  and  the  total 


ELIMINATION     BETWEEN     EQUATIONS.  129 

number  of  these  pieces  was  equal  to  d  ?     How  many  pieces  of  eacli 

kind  were  used. 

hd  —  a        -  a  —  cd 

Am.   — ,  and  — . 

h  —  c  b  —  c 

What  do  these  solutions  become  when  hd  =  a,  and  cd  =.  at  What, 
when  a  >  if/,  and  cd  ^  a?      What,  when  h  =  c?      Explain  these 

results,  especially  the  last. 

C8.  The  sum  of  three  numbers  is  420  :  the  first,  is  to  the  second  as 
G  to  7,  and  the  third  is  just  as  much  greater  than  the  second  as  the 
second  is  greater  than  the  first.     What  are  the  numbers  ? 

Ans.  120,  140,  and  IGO. 

69.  The  sum  of  three  numbers  is  a :  the  first  is  to  the  second  as 
b  to  c,  and  the  second  exceeds  the  first  as  much  as  the  third  exceeds 
the  second.     What  are  the  parts  ? 

.         oc      a        ^  2ab  —  ac 

^^"^  HZ  •  a  ^'^^ —36— 

We  observe  that,  when  quantities  bear  the  above  relation,  the  mean 
is  always  the  third  of  the  whole. 

What  single  hypothesis  will  make  all  three  parts  equal  ?  What  is 
tlie  effect  upon  the  three  part>s  of  making  c  ==  2b?  Explain  this 
result  ? 

70.  At  the  Woman's  Eights  Convention,  held  at  Syracuse,  New 
York,  composed  of  150  delegates,  the  old  maids,  childless-wives,  and 
bedlamites  were  to  each  other  as  the  numbers  5,  7,  and  ?>.  How  many 
were  there  of  each  class  ?  Ans.  50,  70,  and  30. 


ELIMINATION   BETWEEN    SIMULTANEOUS 
EQUATIONS  OF  THE  FIRST  DEGREE. 

207.  Simultaneous  equations  are  those  which  can  be  satisfied  for  the 
same  values  of  the  same  unknown  quantities.  Elimination  can  only  be 
performed  upon  simultaneous  equations.  It  is  a  process  by  which  we 
deduce  from  two  or  more  equations,  containing  two  or  more  unknown 
quantities  a  single  eqiiatiou,  containing  one  unknown  quantity.  The 
single  equation  so  found  is  called  the  final  equation,  and  since  it  con- 
tains but  one  uiiknown  quantity  does  not  lead  to  arbitrary  values. 
I 


130  ELIMINATION    BETWEEN    SIMULTANEOUS 

We  will  first  take  the  simplest  case,  that  of  two  equations  involving 
two  unknown  quantities.  The  elimination  of  one  of  these  unknown 
quantities  may  be  effected  in  four  ways. 

1.  By  Comparison. 

2.  By  Addition  and  Subtraction. 

3.  By  Substitution.  ^ 

4.  By  the  Greatest  Common  Divisor. 


ELIMINATION  BY  COMPARISON. 

208.  Take  the  equations  2y  =  2a;  +  4, 
and  3y  =  6x  —  12. 

From  the  first  we  get        y  =  .x  +  2, 
and  from  the  second,  y  = 'Zx  —  4. 

Now,  since,  by  hypothesis,  the  y  and  x  in  one  equation  are  equal  to 
the  y  and  .r  in  the  other  equation,  we  can  equate  the  values  of  y,  and 
make  the  two  x'&  represent  the  same  thing.  Hence,  a;  +  2  =  2a;  —  4. 
From  which  we  get  a;  =  6.     Solving  the  first  equation  with  respect  to 

y 

X,  we  get  X  =  y  —  2  ;  and  from  the  second  we  get  a;  =  -^  +  2. 
Now,  since  the  .r's  are   equal  by  hypothesis,  we  have  a  right  to 

y 

equate  their  values.     Hence,  y  —  2  =  ^  +2. 

Assuming  that  the  y's  are  equal  in  the  two  members  of  this  equa- 
tion, wo  have  y  =  S.  Hence,  the  values  of  x  and  y  are  x  =  6,  and 
y  =  8.     It  will  be  seen  that  these  values  satisfy  both  equations. 

Elimination  by  comparison  between  two  equations  with  two  unknown 
quantities  consists  in  solving  both  equations  with  reference  to  one  un- 
known quantity,  and  equating  the  values  so  found.  This  eliminates 
the  first  unknown  quantity,  and  gives  a  single  equation,  from  which 
the  value  of  the  second  can  be  found.  The  second  unknown  quantity 
being  eliminated  in  like  manner,  the  value  of  the  first  can  be  found. 

209.  If  there  had  been  three  unknown  quantities,  and  but  two  equa- 
tions, there  would  have  been,  after  the  elimination  of  y,  a  single  equa- 
tion with  two  unknown  quantities,  and  the  values  of  these  two  un- 
known quantities  would,  of  course,  have  been  arbitrary.  If  there  had 
been  two  equations  and  but  one  unknown  quantity,  the  equation  could 
not,  in  general,  be  satisfied  by  the  same  value  for  that  unknown  quan- 


EQUATIONS    OF    THE    FIRST    DEGREE.  131 

tity.  The  equations  .r  +  2  =  a,  and  x  —  2  =  a,  cannot  be  satisfied 
for  the  same  values  of  x. 

We  see,  then,  fJiat  the  nnniher  of  equations  must  he  preciseli/  equal  to 
the  number  of  unknovni  quantities. 

210.  The  equations  have  been  combined  under  the  supposition  that 
they  were  simui^neous.  If  they  are  not  so,  the  hypothesis  has  been 
absurd,  and  the  result  ought  to  indicate  the  absurdity.  Take  thc' 
manifestly  absurd  equations, 

y  =  2./-  +  2, 
and  y  =  2x  —  2. 

Combining,  we  get  2x  —  2x,  or  Ox  =  4.     Hence,  x  ^  ^  ^  cc. 

The  absurdity  of  the  h}^othcsis  is  here  -oointed  out  by  the  symbol 
of  absurdity. 

211.  If  the  equations  arc  the  same,  or  differ  only  in  form,  they  can, 
obviously,  be  satisfied  by  an  indeteriiiinate  number  of  values  for  one  of 
the  unknown  quantities,  provided  that  of  thc  other  is  deduced  after  the 
substitution  of  the  assumed  value  of  the  first. 

Take,  y  =  2x  +  2, 

and  1/  =  2x  +  2. 

Combining,  we  get  0^  =  0,  or  ^  =  — .     Which  indicates  that  y  may 

have  any  value  whatever.  Suppose  we  assume  ^  =  4 ;  this  value  for 
1/,  substituted  in  either  of  the  equations,  will  give  x  =  1;  and  the  two 
values,  y  =  4,  and  x  =  1,  will  satisfy  both  equations.  Assuming 
arbitrary  values  for  t/,  we  get  an  infinite  number  of  values  for  x,  and 
the  solution  becomes  wholly  indeterminate. 


ELIMINATION  BY  ADDITION  AND  SUBTRACTION. 
212.  Resume  the  equations, 

2y  =  2x  +  4, 
and  S>/  =  Qx  —  12, 

Multiply  the  first  equation  by  3,  and  thc  second  in  like  manner  by 
2,  and  we  will  have 

6y  =    6x  +  12 

6y  =  12a:  —  24 

0=    Qx  —  36. 


132  ELIMINATION     BETWEEN     SIMULTANEOUS 

If  tliGse  equations  are  simultaneous,  the  0//  in  the  first  equation  is 
equal  to  the  6y  in  the  second.  And,  since  the  x's  will  also  be  equal, 
the  result  of  the  subtraction  of  the  equations,  member  bj  member,  will 
be  0  =  6x  —  36,  or  x  =  6,  the  same  value  as  before  found. 

Now,  to  eliminate  x  in  order  to  find  the  value  of?/,  the  first  equation 
must  be  multiplied  by  3,  and  the  second  must  remain  as  it  is. 

We  will  get,  Gj/  =  6x  +  12, 

and  3;y  =  6x  —  12, 

and  by  subtraction,  Zy  =  24,  or  y  =  8. 

The  same  value  as  when  we  eliminated  by  comparison. 

It  will  be  seen  that  the  coefficients  of  the  unknown  quantity  to  be 
eliminated  have  bees  made  equal  in  the  two  equations ;  and,  since  they 
had  like  signs,  the  elimination  could  only  be  efiectcd  by  subtraction. 
Had  these  coefficients,  however,  been  affected  with  contrary  signs,  it 
would  have  been  necessary  to  add  the  equations  member  by  member. 
Take  as  an  illustration, 

2y  — x  =  3 

X  — y  =:  —  1,  by  addition,  x  is  eliminated. 
?/  =  2 

To  eliminate  y,  multiply  the  second  equation  by  2,  and  we  get 


2a;  —  2y  =  —  2, 
and  by  addition,  x  z=.\ 

The  two  values  are  then  y  =^  2,  and  x  —  1,  and  these  satisfy  both 
equations. 

Elimination  by  Addition  and  Subtraction  consists  in  making  the 
coefficients  of  the  unknown  quantity  to  be  eliminated  the  same  in  both 
equations,  and  then  subti-acting  the  equations  member  by  member,  if 
these  coefficients  have  like  signs,  or  adding  them  member  by  member 
if  they  have  unlike  signs. 

It  will  be  seen  that  the  method  of  elimination  by  addition  and  sub- 
traction involves  no  fractions,  whilst  the  method  by  comparison,  in 
general,  does  involve  fractions.  To  eliminate  by  comparison  between 
the  equations 

3y  =  a:  +  2, 
and  2y  =  a;  —  2, 

X  -{-  2  X 2 

■will  involve  the  fractions  — ^ — ,  and  — r — . 
o  2 


EQUATIONS     OF    THE    FIRST    DEGREE.  133 


ELIMINATION  BY  SUBSTITUTION. 

213.  This  consists  in  finding  the  value  of  one  of  the  unknown  quan- 
tities in  one  of  tlie  equations  in  terms  of  the  other  unknown  quantity, 
and  substituting  the  value  found  in  the  second  equation,  so  as  to  deter- 
mine the  valueijf  the  second  unknown  quantity  in  known  terms. 

llesume  the  equations, 

and  3y  =  Gj;  —  12. 

From  the  first  we  get  .y  =  x  +  2 ;  and  this  value  of  ?/,  substituted 
in  the  second  equation  (since  the  ^'s  and  x's  are,  by  hypothesis,  the 
same  in  both  equations),  gives  3x  +  6  =  6.t  —  12.  Hence,  x  =  Q. 
In  like  manner,  x,  found  from  the  first  equation,  x  =■  y  — 2,  substi- 
tuted in  the  second,  gives  3?/  =  6  (y  —  2)  —  12,  or  y  =  8.  The 
two  values  are  then  x  =  G,  and  y  =  S,  as  found  by  the  other  two 
methods. 

This  process  will  also,  in  general,  involve  fractions. 


ELIMINATION  BY  THE  GREATEST  COMMON  DIVISOR. 

214.  This  process  consists  in  transposing  all  the  terms  of  the  two 
equations  to  the  first  member,  and  then  dividing  the  polynomials  in  the 
first  member  by  each  other,  as  in  the  method  of  finding  the  greatest 
common  divisor,  until  a  remainder  is  found  which  contains  but  one  un- 
known quantity.  This  remainder,  placed  equal  to  zero,  constitutes  the 
final  equation,  and  the  value  of  one  of  the  unknown  quantities  can  be 
deduced  from  it.  The  value  of  the  other  unknown  quantity  can  be 
found  by  a  similar  process. 

Resume  the  equations, 

2y  =  2x  -f  4, 
and  Sj^  =  Qx  —  12, 

transposing,  2y  —  2x  —  4  =  0, 

and  ?>y  —  Qx  +  12  =  0. 


12 


6y  —  12.r  4-  24 

2y  — 2x  — 4 

6^  — 6x    —12 
—  Gx    —  3G  = 

3            Quotient 
=  0,  or  X  =  6. 

134  ELIMINATION    BETWEEN     SIMULTANEOUS 

To  eliminate  x,  arrange  witli  reference  to  x,  and  we  have 

—  Gx  +  %  +  12  I  —  2x  +  2.y  —  4 

—  6x  +  Cy  —  12  I  ^  3  Quotient. 

—  3y  +  24  =  0,  or  ^  =  S 

It  only  remains  to  be  shown  the  reason  why  the  remainder  is  placed 
equal  to  zero  rather  than  to  10,  or  anything  else. 

Let  A  =  0  represent  the  first  equation  after  it  has  been  prepared  for 
division.  Let  B  =  0  represent  the  second  equation.  Let  Q  represent 
the  quotient  after  the  division  of  A  by  B,  then, 

A  |_B_ 
BQ    Q 


A  — QB  =  0. 


Now,  since  A  is  zero,  and  B  zero,  A  —  QB  is  plainly  zero. 

For  equations  of  the  first  degree  there  will  be  but  one  remainder. 
But  if  the  equations  be  of  a  higher  degree  th-an  the  first,  there  will  be 
two  or  more  remainders.  But  the  same  course  of  reasoning  will  show 
that  each  successive  remainder  must  be  zero.  For  the  first  remainder 
having  been  shown  to  equal  zero,  and  the  divisor  B  also  equal  to  zero, 
the  second  remainder  must  also  be  equal  to  zero.  Let  B'  represent  the 
second  remainder,  and  Q'  the  quotient  resulting  from  the  division  of  B 
byB'. 

Then,  B     |jy_ 

Q^B^   Q^ 
B  —  Q'  B'  ==  0. 

Since  B  =:  0,  and  B'  =  0,  plainly  B  —  Q'B'  must  be  zero. 
The  third  remainder,  the  fourth,  and  so  on,  can,  in  like  manner,  be 
shown  equal  to  zero. 

215.  If  we  combine  equations,  which  are  not  simultaneous,  by  either 
of  the  first  three  methods,  the  absurdity  of  the  hypothesis  is  shown  by 
cc,  the  symbol  of  absurdity  in  the  result.  But  when  we  combine  such 
equations  by  the  fourth  method,  the  absurdity  appears  in  the  final 
equation  having  no  unknown  quantity. 
Take  the  equations 

^  =  2x-i-2, 
and  y  =  2x  +  4. 


EQUATIONS     OF     THE     FIRST     DEGREE.  135 

Combining  by  last  process, 

y  —  2x  —  1  1  y  — 2x  — 4 


—  2a;  — 4 


+  2  =  0,  which,  is  absurd. 
21G.  Though  it  is  usual  to  say  that  the  absurdity  of  combining  equa- 
tions which  are  not  simultaneous  is  shown  by  the  final  equation,  yet 
we  might  retain  the  trace  of  one  of  the  unknown  quantities,  and  then 
we  would  still  have  the  symbol  oo.  In  the  last  example,  we  might 
retain  the  trace  of  y,  and  the  final  equation  would  be  0^  +  2  =  0 ; 
whence  y  =  a;. 

Rcmarlis. 

217.  Of  the  four  methods  of  elimination,  the  last  is  generally  used 
when  the  degree  of  the  second  equation  is  higher  than  the  first,  and  the 
second  method  (by  addition  and  subtraction)  is  preferable  for  simple 
equations,  since  it  does  not  involve  fractions.  Elimination  by  substitu- 
tion is  generally  associated  with  the  other  three  methods  after  the  value 
of  one  unknown  quantity  i«  found ;  this  value  is  generally  substituted 
in  one  of  the  given  equations,  and  we  are  thus  enabled  to  deduce  that 
of  the  other. 

Take  the  equations, 

2y  =  2.r  +  4, 
and  Zij  =  G.c  —  12. 

From  the  first  we  have,  y  =  s  -f  2,  and  from  the  second  y  =  2x 
—  4.  Equating  the  two  values  of  y,  we  get  x  +  2  =  2x  —  4. 
Hence,  x  =  6.  This  value  for  x,  substituted  in  the  first  equation, 
gives  2y  =  16,  or  y  =:  S.  And  substituted  in  the  second,  gives 
3/y  =  24,  or  y  =  8.  Hence,  the  given  equations  are  simultaneous. 
But  if  the  substitution  of  the  value  for  x  in  the  two  equations  gave 
different  values  for  y,  we  would  conclude  that  the  equations  were  not 
simultaneous. 

218.  Examples  in  elimination  heticeen  two  simple  equations  o/ the 
first  degree  invohiiij  two  xtnhioicn  quantities. 

1.  Find  the  values  of  x  and  y  in  the  equations, 

y  =  ax  -{■  h, 

and  y  =  a'x  -\-  h'. 

b  —  V        ^           ah'  —  a'h 
Ans.  X  = ;,  and  y  = r-. 


136  ELIMINATION    BETWEEN    SIMULTANEOUS 

The  hypothesis,  a  =  a',  makes  both  values  infinite.  The  combined 
equations  show  that  when  a  =  a',  a'h  must  equal  ah' ;  that  is,  unequal 
multiples  of  the  same  quantity  must  be  equal,  which  is  absurd.  The 
equations,  in  fact,  represent  two  straight  lines,  and  the  found  values  of 
X  and  y  represent  their  point  of  meeting.  The  hypothesis,  a  =  a',^ 
makes  the  lines  parallel,  and  their  point  of  meeting  is,  of  course,  at  an 
infinite  distance. 

Making  a  =  a'  and  h  =  V,  x  and  y  both  become  g,  or  indetermi- 
nate. In  this  case,  the  two  equations  become  identical,  and  can  be 
satisfied  by  any  values  for  x  and  y,  as  the  symbol  g  indicates.  By  this 
we  mean  that  arbitrary  values  may  be  given  to  either  x  or  y,  and  these 
arbitrary  values  for  one  of  the  unknown  quantities,  taken  in  connection 
with  the  deduced  values  of  the  other,  will  satisfy  both  equations. 

When  a  =  a',  and  b  =  I',  the  two  lines  coincide,  and  their  point 
of  meeting,  being  any  where  on  the  common  line,  is,  of  course,  inde- 
terminate. 

When  h  =  V,  we  will  have  x  =  0,  and  y  ^=l>. 


2.  Combine 

2^  +  3x  =  4, 

and 

y-6.x  =  7. 

Ans.  X  — 

3.  Combine 

2y  -f  3:r  =  0, 

y  — 6x  =  0. 

Ans.  x  :=  0,  and  y  ==  0. 
When  there  is  no  known  term  or  terms  in  the  two  equations  the 
values  of  x  and  y  will  always  be  zero,  since  these  values  will  satisfy 
both  equations. 

4.  Combine 
and 

5.  Combine 
and 

6.  Combine 

and 

Am.  a;  =  8,  and  y  =  30. 


2i/  + 

|— 4^  =  0. 

3y  + 

-5-  =  % 

Ans. 

x  =  1,  and  y 

=  2. 

f  + 

1-1  =  0. 

X  + 

2y  -f  4  =  0. 

Ans.  X  = 

:32|,  y  =  - 

■18  J. 

X 

■4 

X- 

+1+ 

1  +  1^  =  "- 

1  +  y  =  36f . 

EQUATIONS    OF    THE    FIRST    DEGREE.  137 


7.  Combine 

3y  —  3x  +  6  =  0, 

and 

lij  —  lx^  14. 

Ans.  y  =  CO,  and  x  =  co, 

8.  Combine 

^^_!^^+2a'  =  0. 

and 

a'lj        "la'x       Act! 
3    ~    3            3  ■ 

.Irts.  x=  2,  and^  =  £. 

9.  Combine 
and 

i  +  i  +  4  =  0. 

3         '^ 

---—2  =  0. 

Ans.  X  =  --j^i,  and  y  =    |. 
Regard  — ,  or  — ,  as  the  quantity  to  be  eliminated. 

10.  Combine  ii  4.  ii  _4«  =  0. 

y         X 

and  =:  2a. 

y         X 

When  one  of  the  unknown  quantities  is  wanting,  it  may  be  written 
with  a  zero  coefficient. 


11. 

Combine 

y 

=  x  +  2, 

and 

y 

=  2,ory  =  Ox  +  2. 

Ans.  X  =  0,  and  y  =  2. 

12. 

Combine 

x  =  0, 

and 

y  =  7x  +  4. 

Ans.  X  =  0,  and  ^  =  4. 

13. 

Combine 

x+y  =  a 

and 

ax  +  y  =  b. 

Ans.  x  = -,  and  y  = -, 

a  —  1                     a  —  1 

What  do  these  values  become  when  o  =  1  ?     Why  ?     What,  when 
h-^a?     What;,  when  a  =  0  ?     What,  when  h  =  a""  ? 

14.  Combine  ■  —  +  ^  =  c. 

X         b 

and  \-  y  =.  a. 


12* 


_  g  (5  — 1)  _  (c  —  fQ  5 

.  «s.  a;  —     ^^_  _—j-,  an    y  _     ]_  _  ^    • 


138  ELIMINATION    BETWEEN    SIMULTANEOUS 

What  do  these  vahics  become  when  h  =  1?  What,  when  c  =  d? 
What,  wlien  c=  d,  and  &  =  1  ?  What,  when  &  =  0  ?  How  are  the 
results  explained  ? 

15.  Combine  —  -| =  <:•. 

X         y 

.  a'         h'  , 

and  - — 1 =  c 

X        y 

.  oV  —  a'h  ah'  —  ah 

ch'  —  be"  ac'  ■ —  a'c 

Solve  this  example  by  the  four  methods  of  elimination.  When  x  is 
eliminated  by  the  last  method,  the  final  equation  in  y  ought  to  be 
{ad  —  a'c)  y  +  a'h  —  ah'  =  0. 

What  do  the  values  of  x  and  y  become  when  a'h  =  ah' ?  Why? 
What  do  these  values  become  when  hd  =  cU  ?  What,  when  a  =  a', 
h  =  V,  and  c  =  d  ? 


2hc 
and  y  = 


16.  Combine 

y  +  |-y  +  a;  =  c, 

and 

y  —  x  =  c. 

Ans.   X  =  —  -T^ ■ 

26  +  a         -^       26  +  a 

The  final  equation  in  x,  when  y  is  eliminated,  ought  to  be  (26  +  a) 
a;  +  ac  r=  0. 

What  do  these  values  become  when  c  =  0  ?  Why  ?  What  do  they 
become  when  a  =  0  ?     What,  when  6  =  0? 

17.  Combine  -L  +  y  =  2a  (1  +  a). 

and  y  +  2x  —  2«  =  2«l 

Ans.  X  =  a,  and  y  =  1o?. 

Eliminating  by  substitution,  we  get  y  =  2x'^.  Hence,  the  second 
equation  will  give  2a;=  —  2a^  +  2.r  —  2a  =  0,  or  2  (.r^  —  a^)  +  2 
(x  —  a)  =  0,  or  2  (x  —  a)  (x  +  a  +  1)  =  0.  Dividing  out  by  the 
factors  2  (a;  +  a  +  1),  we  have  x  —  a  =  0,  or  x  =  a.  The  same  re- 
sult may  be  obtained  by  the  fourth  method  of  elimination. 


18.  Combine  ^ -f  y  =  26, 


and  _  +  2?/  =  26  +  a. 

X 


EQUATIONS    OF    THE    FIRST     DEGREE.  139 

Combining  by  fourth  method,  we  have 

{rj  —  2h)x  +  y  I  (2y  —  2h  —  a)x  +  a. 
Preparing  for  division  by  multiplying  by 

(2y-2/.-a) 

x(y  — 2i)(2y  — 26  — a)  +  i/(2y  — 26  — a) 

xly  —  2b)  (2i/ ~2b  —  a)  +  ai/  —  2ba  y  —  2b.     Quotient. 


2/  —  2ay  —  26y  +  26a  =  2y  (y  —  a)  —  26  (y  —  a)  =  0.  Remainder. 

or,  {jj  —  a)  {2y  —  26)  =  0,  the  final  equation. 

Divide  out  by  2y  —  26,  and  we  have  y  —  a  =  0,  ory=:a;  and 

this  value  for  y,  substituted  for  cither  of  the  equations,  gives  if-  =  .77 • 

We  might  have  divided  out  by  y  —  a,  and  then  we  would  have  had 
2y  —  26  =  0,  ox  y  =  b;  and  this  value  for  y  would  liave  given  x  =  \. 
The   equations,  then,   admit   of  two   systems    of  values,  y  =  a,  and 

;    and  y  =  b,  and  x  =  1.      Example   17   gives,  also,  a 


19. 

Combine 

and 

20. 

Combine 

and 

26 
second  system,  x  =  —  (a  +  1 ),  y  =  2  («  -f  1/  . 

^+-  =  2, 
x         y 

y  —  x  =  0. 

Ans.  X  =  Q,  and  y  =  g. 

-  + J  -=  2 

X        y 

y  —  x  =  2. 

Am.  x  =  2,  and  y  =  4,  or  .r  =  —  4,  and  y  =  —  2. 

Combining  by  substitution,  we  get  cc*  +  2x  —  8  =  0;  or,  adding 
and  subtracting  unity,  x^  +  2x  +  1  —  9  =  (.f  +  1)^  —  3^  =  (x  +  1 
+  3)  (X  +  1  —  3)  =  0. 

By  suppressing  the  fii-st  factor,  we  get  a;  =  2 ;  and  by  suppressing 
the  second,  we  get  x  =  —  4. 

21.  Combine  yx  —  x  =  6, 

and  X — y  =z  —  2. 

Ans.  X  =  2,  and  ?/  =  4 ;  or  x  =  —  3,  and  y  =  —  1. 

In  this  case,  add  and  subtract  J. 


140  ELIMINATION     BETWEEN     SIMULTANEOUS 

22.  Combine  yx  —  —  =  0. 

y 

and  y  -\-  X  ■=  1. 

Ans.  X  =  1,  and  y  =  1 ;  or  x  =  3,  and  y  =  —  1.  ■ 

23.  Combine  a;  +  ^  +  y  .—  2, 

and  x-\-2by  =  0. 

.                       -46  .  9 

.4«s.  .T  =^   ,  and  y 


6  (a  +  2)'  •^-  1-6  (a +  2)' 

What   do   those   values   become   when    a  =  —  2  ?      What,    when 
6=0? 

24.  Combine         x  -{-  h  —  a  +  ^=  , 

2?/ 

and  2x  -\ ^  =  2a  —  26  +  2. 

a  —  6 

Ans.  X  =  a  —  6;  and  y  ■=  a  —  6. 


25.  Combine 

y  =  ax  -}-  6, 

and 

y  =  2. 

2  - 
-4ns.  X  =  — 

— ,andy  =  2. 

"What  do  these  values  become  when  a  =  0  ?  What,  when  6  =  2';' 
Explain  these  results. 

219.   Elimination  hdioeen  any  numher  of  simidtaneoiis  equations. 

The  same  principles  govern  the  elimination  of  any  number  of  simul- 
taneous equations  as  have  been  shown  to  govern  the  elimination  be- 
tween two  equations  with  two  unknown  quantities.  No  specific  rules 
can  be  given  for  elimination,  because  each  equation  may  contain  all  the 
unknown  quantities ;  or,  a  part  only  of  them  may  contain  all.  It  may 
be  even  that  no  equation,  or  but  one,  contains  all  the  unknown  quan- 
tities. The  main  thing  to  be  observed  is,  to  eliminate  the  same  un- 
known quantity  from  all  the  equations  that  contain  it.  We  will  then 
have  one  unknown  quantity  less  than  before,  and  one  equation  less. 
Continue  the  process  of  elimination,  until  a  single  equation  with  a  single 
unknown  quantity  is  obtained.  If  the  number  of  unknown  quantities 
is  greater  than  the  number  of  equations  it  will  be  impossible  to  obtain 
a  single  equation  with  one  unknown  quantity,  because  the  number  of 
equations  reduced  by  elimination  is  always  eoual  to  the  number  of  un- 


EQUATIONS     OF     THE     FIRST     DEGREE.  141 

known  quantities  reduced.  Two  eliminations  reduce  the  number  of 
unknown  quantities  and  equations  by  two ;  three  eliminations  by  three, 
&c.  It  is  evident,  then,  that  when  the  number  of  unknown  quantities 
exceed  the  number  of  equations,  the  last  equation  obtained  will  contain 
two  or  more  unknown  quantities,  and  will,  consequently,  be  an  inde- 
terminate equation. 

220.  If  the  number  of  equations  exceed  the  number  of  unknown 
quantities,  it  is  plain  that,  before  all  the  equations  have  been  freed 
from  their  unknown  quantities,  we  will  get  a  single  equation  with  but 
one  unknown  quantity.  The  value  of  this  unknown  quantity  then  can 
be  determined,  and  it  may  not  be  such  as  to  satisfy  all  the  equations. 

221.  The  number  of  equations  and  unknown  quantities  must  not 
only  be  equal  to  each  other,  but  the  equations  must  be  different  in  cha- 
racter, not  inform  merely.  The  equations  j/  =  2x  -\-2,  and  2i/  =  4x 
-\-  4,  differ  only  in  form,  and  it  is  impossible  to  eliminate  between 
them. 

EXAMPLES. 

1.  Solve  the  three  equations, 

2^  —  x-{.z=2, 
2i/-\-2x+4z  =  8, 
Bt/  +  13-c  +  3z=  19. 

Ans.  1/  =  1,  X  =^1,  and  z=^\. 

2.  Solve  the  three  equations, 

y  +  x  +  z^^:), 
y  —  :c  +  z=B, 
y  —  X  —  2  =  —  5. 

Am.  y  =  2,  .r  =  3,  and  2  =  4. 

3.  Solve  the  three  equations, 

y-f-cc-fs  =  a+  />  +  r, 
y  —  X  -\-  z  -\-h  —  c  =  «, 
y  —  2x  —  '^z--^a  —  2h  —  3c. 

Ans.  y  z=  a,  X  =hj  and  z  ==  c, 

4.  Solve  the  three  equations, 

x  +  2y  =  4, 

x  -{-  z  =  a, 

y  —  z=l. 

Am  y  =  4  —  (a  +  Z-),  X  r=  2  (a  +  t)  —  4,  and  2  =  4  —  (a  +  26). 


112  ELIMINATION    BETWEEN     SIMULTANEOUS 

5.  Solve  the  tlirec  equations, 

X  —  y  z=.  a  —  h, 
.T  +  2  =  o, 
X  +  y  -\-z  —  c  —  L 
Ans.  y  =  h  —  2,  X  —  a  —  2,  and  z  =  S  -\-  c  —  (a  +  Z>). 

6.  Solve  the  three  equations, 

y  a        ^         z         c 

1  X 

2ax  +  j  —  2a  +  -  +  y  —  z  =  a  —  c  +  l. 

9/  —  2x  +  3z=a  +  Sc  —  2. 

Ans.  y  —  a,  X  =  1,  and  z  ~  c. 

7.  Solve  the  three  equations, 

Ax  +By  +  Cz  +  D  =  0, 
A'x  +  B'y  +  C'z  +  D'  =  0, 
A"x  +  B"y  +  G"z  +  D"  =  0. 

,  _  (AT'^  —  A'V)  D  +  ( A^^C  —  AC^O  ly  +  (AC^  —  AT)  D" 

^ns.  y  —  ^^,_j^„  _  ^„^,^  ^  _^  ^^„^  _  ^g,,^  ^,  _^  ^^^,  _  ^,^^  ^,„ 

_  (B'T^  —  BV")  D  +  (BC^  —  B'V)  D'  +  (S'G  —  BCQ  D'^ 
''^^  ~  (A'B"  — A"B')  C  +  (A"B  —  AB")  C  +  (AB'  — A'B)  C" 

_  (A^^B^  —  A^B^O  D  +  ( AB^^  —  A^^B)  J)'  +  ( A^B  —  ABQ  D^ 
^  ~    (A'B"  —  A"B')  C  +  (A"B  —  AB")  C  +  (AB'  —  A'B)  C"' 

What  do  these  become  when  D,  D',  and  D"  are  all  equal  to  zero  ? 
Why  ?  What  do  they  become  when  either  A,  A',  A",  or  B,  B',  B", 
or  C,  G',  and  C"  all  three  are  zero  ?  Why  ?  What  will  be .  the  effect 
of  making  D,  D',  and  D"  zero,  and  A  =  A'  =  A",  B  =  B'  =  B",  and 

c  .=  C  =  C"  ? 

8.  Solve  the  equations, 

X  +  y  -  a, 

X  +  z  =  b. 

y  +  z  =  c. 

a  -\-  c  —  b            a  +  b  —  c            b  +  c  —  a 
Ans.  y  = ~ ,  X  = ,  z  =  ^ . 

What  hypothesis  will  reduce  these  values  to  zero  ?  What  will  make 
the  first  two  equal  ?     What  all  three  equal  ? 


EQUATIONS    OF    THE    FIRST    DEGREE.  143 

9.  Solve  the  three  equations, 

.X  -f-  ^  +  2^  —  /  =  4, 
x—y  —  z  =  2, 
2x-\-2y  \z=  a, 
Values  indeterminate.     Why  ? 

10.  Solve  the  three  equations, 

X  -^^  y  —  a, 
X         y 

Ix  +  cy  =  0. 
Values  found  from  first  and  second  different  from  those  found  from 
second  and  third. 

11.  Solve  the  equations, 

c  c 

y  +  X  +  z~4:(a  +  b)  =  —n(a  +  h), 
y  —  X  —  2  (a  —  h)  -f  a  +  ::  =  2a  +  h. 

Arts,  y  =  a  —  h,  x  =  h  —  (/,  and  z  ^  a  -\-  b. 

12.  Solve  the  equations, 

ay  +  hx  =  c, 

y  +  x  =  h, 

y  +  J—z  =  a. 

c  —  fr  ah  —  r 

Ans.   y  = :,  X  = ,  and  z  =  b  —  a. 

a  —  b  a  —  b 

What  will  be  the  effect  of  making  h  =  a? 

,  f^    ^      ,  .  y  +  X       V  +  z       y  +  t 

13.  Combine  "^—^  +  '^~-  +  '^^-r—  =  3. 

0  4  0 

X  +  z       x+J  _^  z_+2  ^  3 
5  G  7 

X  —  y  +  t  —  z  =  2, 
x  +  y  +  l  +  t  —  z  =  5. 

Ans.  y  =  1,  X  =  2,  z  =  S,  and  <  =  4. 

14.  Combine  xy  =  00, 

.r+^  =  29. 


A71S.  X  =  0,y  =  10,  and  z  —  20. 


144  ELIMINATION    BETWEKN     SIMULTANEOUS 

15.  Combine  ^  +  ^  +  4  f  4"  =  3G. 

o         2         4         o 

Ans.  .-c  =  18,  y  =  20,  z  ==  40,  and  t  =  60. 

16.  Combine  xy  =  QO, 

X  +  z  =  29. 

a;  — -  =  0. 

y 

Equations  absurd.     Why? 

17.  Combine  a;  =  2z  +  4, 

y  =  3;s  —  6, 

a;  =  2s  — 2. 

Ans.   X  =  ex,  1/  =  ac,  and  z  =  ao. 

18.  Combine  '  x  +  ^  +  z  +  t  =  4, 

x  +  ^  —  z  —  t  =  0, 
I  z  +  t  —  X  —  v/  =  0, 

7/  +  Z  —  X~f=0, 

y  —  z  -{-  X  —  t  +  w  =  1. 
Ans.   X  =  1,  y  =1,  z  =1,  t  =  1,  and  to  =  1. 

19.  Combine  x  +  y  +  z  —  4, 

2x  +  2y  +  2a  =  5 
i/  —  z  =  0. 

Ans.  X  =  cc,  y  =  cc,  and  z  =  oc. 

Equations  evidently  not  simultaneous.     Tlie  combination,  tlien,  is 
absurd,  and  the  result  shows  the  absurdity. 

20.  Combine  x  +  y  +  z  =  S, 

x  —  i/  —  z=  1. 
3a;  +  3y  +  32  =  18. 

Ans.  X  =  oo,  1/  =  cc,  and  z  =  cc. 

The  first  and  third  equations  plainly  conflict      Hence,  the  equations 
are  not  simultaneous. 


EQUATIONS    OF    THE    FIRST    DEGREE.  145 

The  first  and  second  combined  give  x  z=2',  and  this  value  substi- 
tuted in  the  second  equation,  gives  y  +  2;  =  1.  The  values  of  x  and 
y  +  z,  substituted  in  the  third  equation,  give  9  =rr  18. 

But  we  may  have  the  absurdity  shown  by  its  appropriate  symbol  by 
combining  the  first  and  third  equations,  and  retaining  the  trace  of  one 
of  the  unknown  quantities.  Thus,  we  may  have  Ox  =  3,  or  a;  =  00. 
And  so,  likewise,  y  and  z  may  be  shown  to  be  infinite. 

21.  Combine  the  equations, 

—  a;  +  y  +  :;  +  ;  —  t  +  a  =  (a  —  i)  (c  +  (Z  +  e), 
x—y  —  z  +  t  =  a  —  h  +  (h  —  a)  (c  +  d  —  c), 

x+  ^  -Y-.  +-=l(a  —  h), 
c         d         e  ' 

X  y  z  t 

+  -71 r.  +— -. a:  =  4  +  i  — a. 


a  —  b      c  (rt  —  //)        (/  («  —  b)       e  (a  —  h) 

Am.  X  =  (i  —  b,  y  =z{a  —  by,  z  =  (a  —  b^d,  t  =  (a  —  h)e. 
When  will  all  the  values  be  zero?     When  three?     When  the  last, 
only? 

22.  Combine  the  equations, 

2^-3.  ==1, 
2y  +  Sz  =  r, 
y  +  z  =  S, 

ylns.  ^  ^  2,  and  ,~  =:  1. 
IIow  does  it  happen  that  true  solutions  are  found  for  four  equations, 
involving  but  two  unknown  quantities  ? 

23.  Combine         x+y  +  z  +  t-{-u=z  200, 


2   +  3 

+1  + 

t 
"5 

+  ^=50, 

x  +  y- 

-.  =  10, 

x  +  z  — 

/  =  10, 

x  +  t  — 

u  =  10. 

A7is.  a- =  20, 

y-30. 

Z  : 

=  40,  <  =  50, 

and  u  = 

=  60. 

24. 

Combine  the  equations 

X  +  y  + 

z  +  t  + 

11  -. 

=  200, 

X         y 

j  +  i  + 

i-i 

4- 

f  =  50, 

X  +y  —  z 

=  10, 

X+Z~t: 

=  10, 

X  -{-t  —  u 

=  10, 

13 

K 

14G  ELIMINATION    BETWEEN     SIMULTANEOUS 

(A)  y  +  .-  —  «  =  10, 

(B)  ?/  +  <  —  t(  —  20. 

Ans.  Same  as  for  the  la.st  equation. 

In  tlie  last  two  examples  the  new  equations,  marked  (A)  and  (B), 
were  not  incompatible  with  the  others,  and,  therefore,  the  solutions 
found,  from  taking  as  many  equations  as  unknown  quantities,  satisfied 
the  additional  equations.  When,  however,  the  number  of  equations 
exceed  the  nuiubcr  of  unknown  quantities,  and  the  surplus  equations 
are  not  satisfied  by  the  same  values  as  the  other  equations,  the  solu- 
tions will  be  contradictory. 

General  Re.marls. 

2>22,.  If  we  have  m  equations  involving  m  unknown  quantities,  we 
can  find  the  value  of  one  of  these  unknown  quantities  in  terms  of  the 
others,  and  this  value  substituted  in  all  the  other  equations  will  give 
us  m  —  1,  new  equations,  involving  m  —  1  unknown  quantities.  We 
can  now  take  one  of  these  m  —  1  equations,  and  find  the  value  of  a 
second  unknown  quantity  in  terms  of  the  remaining  m  —  2  unknown 
quantities,  and  thus  get  m  —.2  new  equations,  involving  m  —  2  un- 
known quantities.  And,  by  continuing  this  process,  it  is  plain  that, 
after  m  —  1  eliminations,  we  would  get  a  single  equation,  invohing  a 
single  unknown  quantity,  and,  therefore,  would  be  able  to  find  the 
value  of  that  unknown  quantity.  But,  if  we  have  m  equations,  involv- 
ing m  -\-h  unknown  quantities,  we  can  only  make  m  —  1  svibstitutions, 
and  then  we  will  have  a  single  equation,  involving  h  -\-  1  unknown 
quantities.  If  Z*  =  1,  the  final  equation  will  contain  two  unknown 
quantities,  and  bo  of  the  form  x  +  i/  =  10,  an  indeterminate  equation, 
in  which  the  value  of  x  can  only  be  found  by  attributing  arbitrary 
values  to  I/.  HI)  z=  2,  the  final  equation  v,'ill  contain  three  unknown 
quantities,  and  be  of  the  form  of  cc  -f  y  +  s  =  10 ;  in  which  x  can 
only  be  determined  by  giving  arbitrary  values  both  to  y  and  z.  It  is 
plain,  then,  that  when  the  number  of  unknown  quantities  exceed  the 
number  of  equations,  the  elimination  will  lead  to  indetermination. 

223.  If,  on  the  contrary,  we  have  m  -\-  h  equations,  involving  m 
unknown  quantities,  m  of  these  equations  are  suificient  to  determine 
the  m  unknown  quantities.  The  remaining  h  equations  might,  or 
might  not,  be  satisfied  by  the  values  found  from  the  m  equations.  If 
the  I  equations  are  satisfied,  they  are  not  independent  equations ;  if  they 
are  not  satisfied,  they  are  contradictory  equations.     Thus,  x  =  2,  and 


EQUATIONS    OF    THE    FIRST    DEGREE.  147 

2xz=4:  are  satisfied  by  the  same  value  of  x  ;  but  the  second  equation 
is  not  an  independent  equation,  since  it  differs  only  in  form  from  the 
first,  a;  =  2,  and  a;  =  3  are  contradictory  equations.  We  conclude, 
then,  in  general,  that  the  number  of  independent  equations  must  be 
precisely  equal  to  the  number  of  unknown  quantities. 

224.  An  artifice  sometimes  enables  us  to  eliminate  between  a  num- 
ber of  equations  more  readily  than  by  the  usual  direct  process.  As  an 
illustration  take  the  equations, 

X  +  y  =  5, 

X  -{-  z  =  Q, 

z  +y  =  7. 

Let  s  =  x  -{■  7/  -{-  ::.     The  three  equations  then  become 
s  —  z  =  b 

s  —  X  =  7 

Adding,  we  get  3s  —  (x  -|-  y  +  ~)  =  18.     But,  since  x  +  1/  -\-  z 
=  s,  we  have  2s  :=  18,   or  s  =  d.      This,  substituted  in   equations 
marked  A,  gives  2i=  4,  y  =  3,  and  x  =  2. 
Take,  as  a  second  illustration, 

xi/  -\-  xz  =  27, 
7/z-{-y  +  z  =  29, 
xyz  =  GO. 

27 

By  factoring  and  solving  the  first  equation,  we  get  y  +  2  =  — . 

27 

This,  substituted  in  the  second,  gives  ys  +  —  =:  29.      But   from   the 

third  equation  we  find  yz  =  — .     Hence, f-  —  =  29,  from  which 

27 
x  =  ^.  Then,  y  +  ^  =  —  =  9,  and  xyz  =  GO,  or  yz  =  20.  Elimi- 
nating, we  get  z-  —  dz  +  20  ==  0,  or  s«  —  10^  +  25  +  s  —  5  =  0,  or 
(z  —  by  +  z  —  b  =  0,  or  (s  —  5)  (2  —  5  +  1)  =  0.  Dividing  out 
the  second  factor,  we  have  left  z  —  5  =  0,  or  z=zb,  from  which 
y  =  4. 

Take  another  illustration, 

xy  +  xz  =  \2    (1), 

y^-f^=rl08    (2), 

s  —  X  =  8         (3). 


148  ELIMINATION     BETWEEN     SIBIULTANEOUS 

Subtracting  (1)  from  (2),  we  get  y  {z  —  j-)  -\-  z  {z  —  x)  ==  96,  or 
(z  —  X)  {y  +  2:)=r96  (4).  Dividing  (4)  by  (3),  member  by  mem- 
ber, we  got  y  -\-  z  =  12  (5).  Equation  (2)  may  be  put  under  tlie 
form  of  z  (?/  +  z)  =  108.  Dividing  this,  member  by  member,  by  (5), 
we  will  have  2  =  9.  Equation  (1),  put  under  tlie  form  of  x  (y  -^  z) 
=z  12,  divided  by  (5),  gives  x  =  1.  The  value  of  z  substituted  in 
(5),  gives  2/  =  3-     Hence,  x  =  1,  ?/  =  3,  and  z  =  9. 


PROBLEMS  PRODUCING  SIMULTANEOUS  EQUATIONS  OF  THE 
FIRST  DEGREE. 

225.  Many  of  the  problems  already  given  (Arts.  176  and  206)  could 
have  been  solved  as  readily,  or  more  readily,  with  two  unknown  quan- 
tities ;  and  there  are  many  problems  wliich  can  only  be  solved  by  the 
use  of  two  or  more  unknown  quantities. 

We  will  solve  three  of  the  problems  already  given,  to  show  the  man- 
ner of  using  two  unknown  quantities. 

1.  The  sum  of  two  numbers  is  a,  and  their  difference  I,  what  are 
the  numbers  ? 

Let  X  =  greater  number,  and  y  =  smaller.  Then,  by  the  condi- 
tion of  the  problem,  x  +  y  =  a,  and  x  —  y  =zlj.     Eliminating  y  by 

adding  the  two  equations  member  by  member,  we^  get  a;  =  —  -|-  — . 

Eliminating  x  by  subtracting  the   equations  member  by  member,  we 

have  ?/  =  — -.     So,  we  see  that  when  we  know  the  sum  and  diffe- 

2        2 

rence  of  two  quantities,  we  get  the  greater  by  adding  the  half  difference 

to  the  half  sum,  and  the  less  by  subtracting  the  half  difference  from 

the  half  sum. 

Thus,  the  sum  of  two  numbers  is  20,  and  their  difference  10 ;  the 

greater  is,  by  the  formula,  15,  and  the  smaller  5. 

2.  A  fox  is  125  of  his  own  leaps  ahead  of  a  greyhound,  and  makes 
6  leaps  to  the  greyhound's  5 ;  but  two  leaps  of  the  greyhound  are 
equivalent  to  3  leaps  of  the  fox.  How  many  leaps  will  the  fox  take 
before  he  is  overtaken  by  the  greyhound  ? 

Let  X  =  distance  passed  over  by  the  fox,  counted  in  terms  of  his 
own  leaps.  Let  y  =  distance  passed  by  the  greyhound.  Then  these 
distances  would  be  proportional  to  the  relative  number  of  leaps  taken 
by  the  fox  and  gi'cyhound,  if  their  leaps  were  equal  in  length.     And 


EQUATIONS    OF    THE     FIRST     DEGREE.  149 

we  would  have  x,  y  :  :  6  :  5.  But  since  each  leap  of  the  greyhound  is 
equivalent  to  |  leaps  of  the  fox,  the  5  leaps  of  the  greyhound  are  equi- 
valent to  5  .  I  =  ij^  leaps  of  the  fox.  Hence,  x  :  y  :  :  Q  :  L^,  and 
the  distance,  y,  passed  by  the  greyhound,  will  be  expressed  in  terms 

5x 
of  the  leaps  of  the  fox.     From  the  proportion  we  get  y  =  —,  and  from 

the  condition  of  the  problem,  y  —  x  =  125.  Combining,  we  get 
X  =  500,  and  y  =  625  leaps  of  the  fox. 

It  will  be  readily  seen  that  this  solution  has  some  advantages  over 
that  with  one  unknown  quantity. 

3.  Two  couriers  start  from  different  points  on  the  same  road,  and 
travel  in  the  same  direction.  The  forward  courier  travels  at  the  rate 
of  b  miles  per  hour,  and  the  rear  courier  at  the  rate  of  a  miles  per 
hour.  They  were  separated  by  a  distance  of  m  miles  at  starting. 
How  long  will  it  be  until  they  come  together  ? 

Let  X  =  distance  travelled  by  forward  courier,  y  =  distance  tra- 
velled by  the  rear  courier.  Then,  since  the  distances  travelled  must  be 
proportional  to  the  rates  of  travel,  we  will  have  x  :  y  :  :  h  :  a,  or  by 
=  ax.      From  the  conditions  of  the  problem,  we  have  y  —  xz=m; 

combining,  we  get  x  = r,  and  y  = -,.      These  di.stances,  di- 

u  —  (/  u  —  o 

vided  by  the  rate  of  travel,  will,  of  course,  give  the  time  elapsed  before 

their  junction.     Both  expressions  give  for  the  time, as  found, 

a  —  b 

when  one  unknown  quantity  was  used.     This  solution,  compared  with 

the  preceding,  shows  how  much  the  use  of  two  unknown  quantities 

has  shortened  the  work. 

4.  A  carpenter  wishes  to  saw  a  piece  of  timber,  20  feet  long,  into 
two  parts,  so  that  one  of  them  shall  be  but  two-thirds  as  loii"-  as  the 
other.     "Where  must  he  place  his  saw  ? 

Ans.  At  12  feet,  or  8  feet  from  the  end. 

5.  A  carpenter  wishes  to  saw  a  piece  of  timber  7n  feet  long,  into  two 
such  parts,  that  one  shall  be  the  —  part  of  the  other.     Where  must  he 

place  the  saw  ? 

.  .  ^      b)}i  am 

Alls.  At  -— —  feet,  or feet  from  the  end. 

b  +  a  b  +  a 

What  values  must  b  and  a  have  to  make  the  solutions  in  examples 
4  and  5  the  same  ?     When  will  the  parts  be  equal  ? 
13* 


150  ELIMINATION    BETWEEN     SIMULTANEOUS 

6.  A  colonel  wishes  to  divide  his  regiment  of  800  men,  and  10  com- 
panies, in  such  a  manner  that  the  two  flank  companies  shall  contain 
each  one-third  more  men  than  each  of  the  central  companies.  What 
must  be  the  number  of  men  in  each  flank  company,  and  in  each  cen- 
tral company  ? 

^??s.  100  men  in  each  flank  company,  and  75  in  each  central  com- 
pany. 

7.  A  colonel  wishes  to  divide  his  regiment,  composed  of  a  men,  and 
b  companies,  in  such  a  manner  that  the  two  flank  companies  shall  each 

contain  —  more  men  than  each  central  company.     What  must  be  the 

composition  of  the  flank  and  central  companies  ? 

Ans.  Central  companies,  each -^  ;  flank  companies,  each-v- 7^ ■ 

be  +  Z  be  +  2 

What  values  must  a,  b,  and  c  have  to  make  this  solution  the  same 

as  the  last  ?     What  do  the  expressions  become  when  c  =  0  ?     What 

do  the  central  companies  become  in  that  case  ? 

8.  A  man  becoming  insolvent,  leaves  $4000  to  his  two  creditors  :  to 
one  of  whom  he  owes  $8000,  and  to  the  other  $6000.  What  share 
ought  each  to  have  out  of  the  $4000  ? 

Ans.  The  first  $2285|,  the  second  |;i714f . 

9.  A  man  becoming  insolvent,  leaves  a  dollars  to  his  two  creditors. 

To  one  he  owes  b  dollars,  and  to  the  other  c  dollars.     What  share 

ought  each  to  have  ?  .  ab  ^  ac 

A71S.  X  = ,  and  V  == . 

When  will  these  values  becomes  equal  ?  When  the  first  double  of 
the  second  ?  What  do  they  become  when  he  leaves  nothing  ?  What, 
when  he  leaves  enough  to  pay  his  creditors  ? 

10.  A  planter  hired  a  negro-man  at  the  rate  of  $100  per  annum,  and 
his  clothing.  At  the  end  of  8  months  the  master  of  the  slave  took 
him  home,  and  received  $75  in  cash,  and  no  clothing.  What  was  the 
clothing  valued  at  ?  Ans.  $12  J. 

Verify  this  result. 

11.  A  planter  hired  a  negro-man  at  the   rate   of   a  dollars    for  c 

months,  and  was  also  to  give  the  negro  a  year's  supply  of  clothing. 

At  the  end  of  b  months  the  negro  was  taken  away,  and  the  planter 

paid  m  dollars  in  cash,  and  gave  no  clothing.     What  was  the  clothing 

valued  at  ?  .  cm  —  ab 

Ans.  y  = r . 


EQUATIONS    OF    THE    FIRST    DEGREE.  151 

When  will  2/  =  0  ?     When  -will  it  be  negative  ?     Wlieu  iufiuite  ? 
The  zero  solution  can  be   explained  most  satisfactorily  by  placing 

cm  =  ab  under  the  form  —  =— .     Let  the  pupil  make  a  problem  to 

explain  the  negative  solution. 

12.  A  planter  has  500  acres  of  cultivated  land,  which  he  wishes  to 
plant  in  such  a  manner  that  he  may  have  twice  as  much  cotton  as  corn, 
and  three  times  as  much  corn  as  small  grain.  What  division  of  his 
land  must  he  make  ? 

Ans.  50  acres  of  small  grain;  150  acres  of  corn;  and  300  acres  of 
cotton. 

13.  A  planter  has  m  acres  of  land  and  wishes  to  cultivate  it  all  so 
that  he  may  have  a  times  as  much  cotton  as  corn,  and  b  times  as  much 
corn  as  small  grain.     How  much  of  each  kind  must  he  have  ? 

Ans.  z  =  q , J  acres  of  small  grain,  y  =  z^ , — ; — j  acres  01 

1  -i-  b  +  ab  "=        '  -^        1  +  b  +  ab 

corn,  and  x  = , 5 r  acres  of  cotton. 

'  1  +b  +  ab 

What  suppositions  upon  a,  b,  and  vi  will  make  this  solution  the 
same  as  the  last  ?  What  two  suppositions  will  make  the  three  divisions 
of  land  equal  ?  What  will  be  the  effect  of  increasing  b  upon  the  values 
of  z,  y,  and  x?  What  of  increasing  or  decreasing  m?  Why  does  m 
enter  into  the  three  numerators  ? 

14.  In  the  year  1G92,  the  people  of  Massachusetts  executed,  impri- 
soned, or  privately  persecuted  409  persons,  of  both  sexes,  and  all  ages, 
for  the  alleged  crime  of  witchcraft.  Of  these,  twice  as  many  were  pri- 
vately persecuted  as  were  imprisoned,  and  7j|  times  as  many  more 
were  imprisoned  than  were  executed.  Ile(][uircd  the  number  of  suffer- 
ers of  each  kind  ? 

Ans.  19  executed,  150  imprisoned,  and  300  privately  persecuted. 

15.  A  planter  has  $2500  to  expend  in  the  purchase  of  30  head  of 
horses  and  mules.  He  wishes  his  horses  all  to  be  equal  in  value,  and 
his  mviles  all  to  be  equal  in  value,  but  each  mule  to  be  one-fourth  less 
valuable  than  each  horse,  and  the  number  of  mules  to  be  twice  as 
great  as  the  number  of  horses.  What  must  be  the  price  of  each  horse, 
and  of  each  mule  ?  Ans.  Each  horse  $100,  each  mule  $75. 

16.  A  planter  has  a  dollars  to  expend  in  the  purchase  of  b  head  of 
horses  and  umles.    He  wishes  to  have  c  times  as  manv  mules  as  horses, 


152  ELIMINATION     BETWEEN     SIMULTANEOUS 

but  cacli  nnilc  to  be  —  times  less  valuable  than   each  horse.      What 
a 

must  be  the  priee  of  each  mule,  and  of  eacli  horse  ? 

Ans.  Each  horse  ,  ^  ,    — , — —- :   each  mule  ~ — , ^ -^. 

b  {d  -f-  cd  —  c)  '  b  (jci  +  cd  —  c) 

What  values  must  be  given  to  a,  b,  c,  and  d  to  make  this  solution 
the  same  as  the  last  ?  IIow  can  the  solution  be  verified  ?  What  will 
be  the  effect  of  making  (Z  =  1  ?  Why  ?  What  is  the  effect  of  making 
c  =  1.     The  expressions  for  the  entire  cost  of  the  horses  and  mules 

and  -3—^ — - — -- .     What  is  the  effect  of  making  f  =  0 


d  +  cd  —  c  d  +  cd 

upon  these  values  ?  What  of  making  a  =  0  ?  How  must  the  expres- 
sions for  the  price  of  eachi  liorse  and  mule  be  written  to  show  that  the 
former  decreases  with  the  increase  of  d,  and  that  the  latter  increases 
with  the  increase  of  d  ? 

17.  The  sum  of  two  digits  is  6,  and  the  second  digit  is  double  the 
first.     What  is  the  number  made  up  of  these  two  digits  ? 

Ans.  24. 
The  digits  are  the  individual  figures  making  up  a  number.     In  this 
example  2  is  the  first  digit,  and  4  the  second. 

18.  The  sum   of  two  digits  is  a,  and  the  second  digit  is  b  times 

greater  than  the  first.     What  are  the  digits,  and  what  is  the  number  ? 

,  a  ah  ^  10a  -{-  ab 

Ans.  X  =  — -^,  2/  =  ^^1 ;  number  =  -y^^^- 

The  number  24  is  made  up  of  the  digits  2  and  4,  the  number  42  of 
those  digits  in  reverse  order.  To  express  that,  18  added  to  the  first 
number  would  reverse  the  digits;  we  represent  by  x  and  y  the  digits 
in  the  first  number.     Then,  lOx  +  y  +  18  =zlGy  -\-  x. 

19.  A  number  is  made  up  of  two  digits,  and  the  first  is  double  the 
second.  If  27  be  taken  from  the  number,  the  digits  will  be  reversed. 
What  is  the  number?  Ans.   63. 

lOx  +  y  —  27  =  IQy  +  X,  and  x  =  2y. 

20.  A  number  is  made  up  of  two  digits,  and  the  first  is  a  times  as 
great  as  the  second.  And  if  b  be  taken  from  the  number,  the  digits 
will  be  reversed.     What  is  the  number,  and  what  are  the  digits  ? 

Ans.  First  digit  ^'^^  ;  second  g^^^-    Number,  f^^, 
(10a  +  I)  b 


EQUATIONS    OF    THE    FIRST    DEGREE.  153 

What  will  these  results  become  when  a  =  1  ?  Does  the  equation 
of  the  problem  indicate  the  absurdity  '(  What  will  the  number  become 
when  i  =  a  —  1  ?  What,  when  a  <^\'i  How  is  the  negative  solu- 
tion explained  ? 

21.  A  number  is  made  up  of  three  digits,  whose  sum  is  equal  to  10. 
The  first  digit  is  double  the  second,  and  the  second  triple  the  third. 
What  is  the  number  ?  Ans.   G31. 

22.  A  number  is  made  up  of  three  digits,  whose  sum  is  equal  to  a. 
The  first  is  h  times  as  great  as  the  second,  and  the  second  c  times  aa 
great  as  the  third.     AVhat  are  the  digits,  and  what  is  the  number? 

,  a  ac  acb 

_  100oc6  +  lOac  +  a 
he  -\-c-\-\         ■ 

What  will  be  the  effect  of  increasing  c  upou  the  three  digits  ?  What 
of  increasing  h  ?  What  upon  the  digits  and  number  of  making 
6  =  0? 

23.  A  number  is  made  up  of  three  digits,  whose  sum  is  equal  to  10. 
The  first  digit  is  double  the  second;  and  if  495  be  taken  from  the 
number,  the  digits  will  be  reversed.     What  is  the  number  ? 

Ans.  631, 

24.  The  sum  of  the  three  digits  which  make  up  a  number  is  equal 
to  m.  The  first  digit  is  a  times  as  great  as  the  second,  and  if  h  be 
taken  from  the  number  the  digits  will  be  reversed.  What  are  the 
digits,  aqd  what  is  the  number  ? 

mam  —  6  (a  -f  1)  99ni  +  h  (99m  -\- h)  a 

(2a  +  1)  99      '  y~  (2a  +  1)  99'  (2a  +  1)  99' 

100a  (90??i  -f  h)  +  10  (99m  +  J)  +  99am  —  b(a  +  1) 

and  number  =  — ^ ^2,  +  1)  99 

25.  A  gentleman  in  Piichmond  expressed  a  willingness  to  liberate 
his  slave,  valued  at  SlOOO,  upon  the  receipt  of  that  sum  from  charitable 
persons.     He  received  contributions  from  24  persons;  and  of  these 


What  was  the  entire  amount  given  by  the  latter  ? 

A71S.  $50  by  the  former;  6950  by  the  latter. 

26.  If  7i  be  taken  from  the  numerator  and  denominator  of  a  certain 


15i  ELIMINATION     BETWEEN     SIMULTANEOUS 

fraction,  its  value  will  be  doubled;  but,  if  6f  be  taken  from  the  nume- 
rator and  denominator,  its  value  will  be  trebled.    What  is  the  fraction? 

Ans.   |. 

Wbat  fraction  is  that  from  wbich,  if  a  be  taken  from  both  its  terms, 
the  value  will  be  doubled ;  and  if  b  be  taken  from  both  its  terms,  the 
value  will  be  trebled  ? 

2a  —  h  ah 


Ans. 


4a  —  06'       4a  —  ob 


ah 


2a  —  b 
The  first  result  gives  the  reduced  fraction,  the  second  (which  is 
identical  with  it)  the  fraction  in  which  the  substitution  must  be  made. 
We  will  illustrate  by  a  problem. 

27.  Find  a  fraction,  such,  that  if  5  be  taken  from  both  its  terms, 
the  value  of  the  fraction  will  be  doubled ;  but,  if  4  be  taken  from  both 
its  terms,  the  value  will  be  trebled. 

Ans.    f,or^ 

9" 

The  subtraction  of  5  from  the  numerator  and  denominator  of  the 
second  fraction  will  give  |,  and  the  subtraction  of  4,  in  like  manner, 
will  give  |.  But  these  results  could  not  be  obtained  by  operating  on 
the  first  fraction.     The  reason  of  the  difference  is  obvious. 

28.  If  A  and  C  can  do  a  piece  of  work  in  3  days,  B  and  C  together 
in  7  days,  and  A  and  B  together  in  3|,  in  what  time  can  each  per- 
son do  the  work  alone  ? 

Ans.  A  in  4i  days;  B  in  21  days,  and  C  in  lOJ. 

29.  A  and  C  can  do  a  piece  of  work  in  a  days,  B  and  C  can  do  the 
same  in  b  days,  and  A  and  B  the  same  in  c  days.  In  how  many  days 
can  each  one,  alone,  do  the  work  ? 

Ans.   A  in  - — ■ days,  B  in  • --^— days,  C  in 

be  +  a  {b  —  c)      -^  '  ac  -\-  b  (^a  —  c)      "^  ' 

2^7?;^ 


be  —  a  (6  —  (•) 

What  do  these  values  become  when  a  =zc?  What,  when  b  —  r? 
What,  when  a  =  b  =  c?  Suppose  a  {b  —  f )  ]>  be,  will  c  be  a  co- 
operator,  or  a  draw-back  ? 


EQUATIONS     OF     THE     FIRST     DEGREE.  155 

30.  If  A  can  do  a  piece  of  work  in  4i  days,  B  in  21  days,  and  C  in 
10^  days,  how  long  will  it  take  them  all,  working  together,  to  do  it  ? 

A71S.  X  =  21  days. 
Solved  by  a  single  equation, 

31.  If  A  can  do  a  piece  of  work  in  a  days,  B  in  Z»  days,  C  in  c  days, 

how  long  will  it  take  them  all,  working  together,  to  perform  it  ? 

.  ale 

Ans.  X  =  — days. 

ab  +  ac  +  be      "^ 

If  &  =  0,  then,  X  ^0,  an  absurd  result.  But,  by  going  back  to 
the  equation  of  the  problem,  one  of  the  terms  is  infinite,  aud  the  ab- 
surdity appears  under  its  appropriate  symbol. 

32.  A  farmer  has  a  piece  of  land,  worth  $800,  and  two  negroes. 
The  first  negro  and  land  together  are  worth  three  times  as  much  as 
the  second  negro,  and  the  second  negro  and  land  together  arc  worth 
just  as  much  as  the  first  negro.     What  is  the  worth  of  the  negroes  ? 

Alls.  First,  81G00 ;  second,  $800. 

33.  A  former  has  a  tract  of  land,  worth  a  dollars,  and  two  slaves. 
The  first  slave  and  the  land  together  are  worth  b  times  as  much  as  tho 
second  slave,  and  the  second  slave  and  land  together  are  just  equal  in 
value  to  the  first  slave.     What  is  the  value  of  the  slaves  ? 

Ans.  First,  — ^^ —  dollars;  second,      "        dollars. 

o  —  1  b  —  1 

What  do  these  values  become  when  b  =  1? 

34.  A  has  a  number  of  five  and  three-cent  pieces  in  his  pocket ;  B 
wishes  to  get  24  of  them,  and  gives  A  one  dollar.  How  many  pieces 
of  each  kind  must  he  get  ? 

Alls.   14  five-cent  pieces,  aud  10  three-cent  pieces. 

35.  A  has  two  kinds  of  pieces  of  money  in  his  pocket ;  the  first 

worth  a  cents  each ;  and  the  second  b  cents  each.     B  wishes  to  get  m 

of  them,  and  gives  A  c  cents.     How  many  pieces  of  each  kind  must 

he  get  ? 

.  am  —  e  ,  ,  .    ,        ,  <"  —  bm  ^       ,  .    , 

Ans.  X  = 5-  second  kind,  and  y  = first  kind. 

a  —  b  -^        a  —  b 

Suppose  a  =  b,  aud  explain  the  absurdity  of  the  solution.  What  is 
the  meaning  of  the  solution  when  am  =  c?  What,  when  am^<^  c? 
What,  when  bm  ^  c  ? 

36.  A  has  two  kinds  of  money.     It  takes  8  pieces  of  the  first  kind, 


156  ELIMINATION     BETWEEN     SIMULTANEOUS 

and  83i  pieces  of  the  second  kind  to  be  worth  a  dollar.     3?  offers  him 
a  dollar  for  27  pieces.     How  many  pieces  of  each  kind  must  he  get? 
Ans.  2  of  the  first,  and  25  of  the  second  kind. 

37.  A  has  two  kinds  of  money.  It  takes  a  pieces  of  the  first  kind, 
and  b  pieces  of  the  second  kind  to  make  a  dollar.  A  dollar  is  offered 
for  c  pieces.     How  many  of  each  kind  must  be  given  ? 

Ans.   — r of  the  first  kind,  and  — y of  the  second  kind. 

b  —  a  b  —  a 

Explain  the  solution  when  b  =  a.     When  b  =  c.     When  a  =  0, 

or  c. 

38.  A  certain  person  has  a  certain  sum  of  money,  which  he  jjlaced 
out  at  a  certain  interest.  A  second  person  has  a  less  sum  by  $16661, 
which  he  puts  out  at  one  per  cent,  more  interest  than  the  first  got,  and 
receives  the  same  income  as  the  first.  A  third  person  has  a  less  capi- 
tal than  the  first  by  $2857^,  but  invests  it  two  per  cent,  more  advan- 
tageously, and  also  receives  the  same  income.  What  are  the  three 
sums  at  interest,  and  what  the  respective  rates  of  interest  of  each  ? 

Capitals,  $10,000,  $8333^,  and  $7142f . 
Rates  of  interest,  5  per  cent.,  6  per  cent.,  and  7  per  cent. 

39.  A  gentleman  invests  a  certain  capital  at  a  certain  rate  of  inte- 
rest. A  second  gentleman  has  a  less  capital  by  a  dollars,  but,  by  in- 
vesting it  at  one  per  cent,  more  advantageously  he  derives  as  much 
income  as  the  first.  A  third  gentleman  has  a  less  capital  than  the  first 
by  b  dollars,  but,  by  investing  it  at  two  per  cent,  more  advantageously, 
he  also  receives  the  same  income  as  the  first.  Required  the  three  capi- 
tals, and  the  three  rates  of  interest. 

Ans.  Capitals,  $- Tj^—^i r~)^-r, r"- 

^        ^     2a  —  b         Za  —  b  la  —  b 

2  (b  —  a)  h  ^      2a 

Rates  of  mtercst,  — r^ ^,     ^ r,  and =-. 

'    2a  —  b         2a  —  b  La  —  b 

Verify  these  results.  Discuss  them  when  b  =  a.  When  b  =  2a, 
When  b  =  0.     When  a  =  0,  &c. 

One  hundred  parts  of  gunpowder  are  composed  of  the  following  ma- 
terials in  the  following  proportions  : 

For  -war.                     For  hunting.  For  mining. 

Nitre             75                            78  65 

Charcoal         12 1                            12  15 

Sulphur         12,^                          10  20 

100  100  100 


EQUATIONS    OF    THE    FIRST    DEGREE.  157 

40.  At  the  beginning  of  the  Mexican  war,  the  proprietor  of  the  Du- 
pont  JNIills  wished  to  work  up  the  materials  of  his  powder  for  hunting 
and  mining,  and  make  war  powder  out  of  it.  He  removed  the  sulphur 
by  sublimation,  and  then  wished  to  ascertain  what  proportion  to  take 
of  the  remaining  charcoal  and  nitre  in  the  two  specimens.  What  pro- 
portion ought  he  to  have  taken  ? 

Ans.  The  proportion  of  the  hunting  materials  to  that  of  the  mining, 

as  i|g  is  to  2^5 ;  or  as  125  is  to  30 ;  or  as  25  to  6. 

Then,  calling  25x  the  amount  of  hunting  material  in  the  nitre,  we 

"5  X  75 
have  25x  +  ^-^^  =  "5,  or  x  =  |-|.       There  ought  then  to  be         '  — 

G  X  75 
parts  of  the  hunting  material,  and  — -j—  parts  of  the  mining  material 

to  give  75  parts  of  nitre  in  the  war  mixture.     A  similar  relation  can  be 
obtained  for  the  proportion  of  charcoal. 

41.  The  sura  of  four  numbers  is  107.  The  first,  increased  by  8,  the 
second,  increased  by  4,  the  third,  divided  by  2,  and  the  fourth,  multi- 
plied by  4,  will  all  give  equal  results.     "What  are  the  numbers? 

Ah!^.  20,  24,  50,  and  7. 

42.  Tlie  sum  of  4  numbers  is  a.  The  first,  increa.sed  by  b,  the 
second,  increased  by  c,  the  third,  divided  by  (f,  and  the  fourth,  mul- 
tiplied by/,  will  all  give  equal  results.     AVhat  are  the  numbers? 

(a  _  c)  /—  (1  -f  //)  h   (c  +  a)/+  (.//--f  l)c  — (l+/0?> 
^"'-        '\2+d)/+l         '  (2-1-  J)/+l 

\2/i  +  (a-c^/\d  2h+a-c 

(2  +  c/)/+l  '  ^(2-|-cO/+r 
Verify  these  results  by  addition.  Show  that  their  sum  is  equal  to  a. 
Verify  them  by  adding  h  to  the  first,  c  to  the  second,  multiplying  the 
fourth  by/,  and  dividing  the  third  by  J.  What  single  supposition  will 
make  tlie  first  and  second  equal  to  each  other  ?  What  single  supposi- 
tion will  make  the  third  and  fourth  equal  ?  What  will  make  the  first 
part  zero  ?    What  effect  will  this  hypothesis  have  upon  the  other  parts  ? 

43.  Four  persons  owe  a  certain  sum  of  money:  of  which  the  first  is 
to  pay  one-third,  the  second  one-fourth,  the  third  one-fifth,  and  the 
fourth  one-sixth.  After  paying  a  portion  of  the  money,  there  is  still 
a  deficiency  of  $36.     What  portion  of  it  has  each  to  pay  ? 

Ans.  The  first,  Sl2|of ;  the  second,  $dj%\;  the  third,  67yVT;  ^^^ 
fourth,  SG/,\. 

Let  60.r  =  the  proportion  of  the  first. 
14 


158  EQUATIONS    OF    THE    FIRST    DEGREE. 

•i4.  Four  persons  owe  a  debt  of  a  dollars :  of  wliicli  the  first  is  to 
pay  the  — tli  part;  the  second,  tlic  — th  part;  the  third,  the  — th  part; 

and  the  fourth,  the  —  th  part.     What  has  each  to  pay? 

Ans.  The  first,  -^-^ — ,    ,^  — , — ; r-  :  the  second, 

'  cd  {c  -\-h)  +  be  (e  +  d)'  ' 

aide  ^,      ^,  .   ,  ahce 

the  third,   — ^— — — — ;    the 


ed  (c  +  6)  +  he  (e  +  (/) '  '   ed  (c  +  6)  +  be  {e  +  d)' 

fourth, 


(c  +  6)    +  be  (e  +  d) 
What  hypothesis  will  make  the  first  two  results  equal  ?     What  the 
second   two  ?     What  all  four  ?     What  will  be  the   effect  of  making 
either  h,  c,  d,  or  e  equal  to  zero.     How  is  this  explained  ? 

45.  The  denominator  of  one  fraction  is  4,  and  of  a  second  fraction, 
8 ;  and  the  numerator  of  the  second  fraction  is  4  times  as  great  as  the 
numer^^or  of  the  first.  The  two  fractions  and  their  greatest  common 
divisor,  added  together,  are  equal  to  3.     What  are  the  numbers  ? 

Ans.   i,    igi,     and  |. 

46.  The  denominator  of  one  fraction  is  h,  and  that  of  a  second  frac- 
tion is  c;  the  numerator  of  the  second  fraction  is  m  times  greater  than 
that  of  the  first,  and  the  sum  of  the  two  fractions  is  equal  to  the  least 
common  multiple  of  their  denominators.     What  are  the  fractions  ? 

Ans.   -,  and 


c  +  mb  c  +  mb 

b  ~c 

47.  A  1000  cubic  inches  of  bronze  were  found  to  weigh  5100  ounces. 
A  cubic  inch  of  copper  weighs  b\  ounces,  and  a  cubic  inch  of  tin 
weighs  4i  ounces.  What  was  the  proportion  of  copper  and  tin  in  the 
composition  of  bronze  ? 

Ans.  The  copper  to  the  tin  as  85  to  15. 

48.  Some  inspectors  of  cannon  weighed  in  cubic  inches  of  bronze, 

and  found  the  weight  to  be  to  ounces.     A  cubic  inch  of  copper  weighs 

h  ounces,  and  a  cubic  inch  of  tin  weighs  c  ounces.     How  much  copper, 

and  how  much  tin  was  in  the  composition  ? 

to  —  mc  „  ^mb  —  w  . 

Ans.   — ounces  of  copper,  and  — ounces  oi  tin. 

b  —  c  b  —  c 

What  will  these  values  become  when  b  z=  c  ?     Going  back  to  the 


VANISHING    FRACTIONS.  159 

equation  of  the  problem,  -what  will  b  =^c  show  in  regard  to  tv  and  mb  ? 
Suppose  c  =  0,  what  will  the  results  show  ?  Suppose  ^o  =i  0,  what 
will  both  solutions  become  ? 


VANISHING    FRACTIONS. 

The  symbol  g  has  been  interpreted  to  sitrnify  iudctermiiiation,  and 

this  is  the  true  interpretation  for  solutions  of  equations  of  the  first 

degree,  when  the  symbol  proceeds  from  two  suppositions,  made  either 

upon  the  values  found,  or  upon  the  equation  of  the  problem.     But  the 

symbol  may  arise  from  a  single  hypothesis,  and  then  it  always  indicates, 

not  indctermination,  but  the  existence  of  a  common  factor. 

x-a y^ 

Take  the  expression, —,  which  becomes  g  whcn=j: — y.     But, 

X       y 

by  factoring  the  numerator,  we  have —  = ^-'-A- — -^'2  __ 

^  ^  x  —  y  x  —  y 

X  +  y  ^=  2y,  when  x  =zy.  The  true  value  of  g  in  the  present  in- 
stance is  2y,  as  shown  by  removing  the  common  factor,  x  —  ?/.    Again, 

.        X  —  y         0       ,  T-,        ■'■  —  V 

take   the   expression -,  =  —    when    x  =  y.       But   —, ~  = 

^  .(,'  —  if         0  "^  x^  —  y^ 

- — ^7 — ; = =  -r  when  x  :=  ?/.     And  the  true  value  of 

(x  —  y){x-i-y)       x-^y       -ly 

the  vanishing  fraction  is  again  shown  to  be  finite.     But,  take  -^- —. 

(•*:  —  y) 

=  J-  when  x  =  y.    Factoring,  we  get  -, — ^ .  = =  — , 

or  Qo  when  x  =  y.    And  the  true  value  of  the  vanishing  fraction  in  the 

(x yf         0 

present  instance  is  infinity.     Take  again, ^—  =  —  when  x  =  y. 

Factoring,  we  get  — — — — ^^^ —  z=i  x  —  y  =  0  when  x  =  y.     So, 

we  see,  that  when  there  is  a  common  factor  existing  between  the  nu- 
merator and  denominator  of  a  fraction,  which  factor  has  become  zero^ 
the  fraction  may  have  either  of  three  values,  finity,  infinity,  or  zero. 

To  show  it  more  generally,  take  the  expression  -~ =  —  when 

Q  {x  —  (/)"        0 

x  =  a.      There  may  be  three  cases,  m  may  be  =  n,  <^  n,  or  ^  n. 


160  VANISHING     FRACTIONS. 

p  (x a")™ 

In  the  first  case,  when  m==in,  the  fraction  becomes  j—^^ r— = 

Q  (x  —  a)™ 

P 

— ,  a  finite  quantity.     In  the  second  case,  divided  by  (.c  —  o)"",  the 

P 

fraction  becomes  /r- ^ ,  which  is  infinite,  when  x  =z  a.     In  the 

third  case,  when  m  ^  n,  dividing  by  (x  —  o)",  the  fraction  becomes 

Y  (x a)""" 

— ^^ — =  0  when  X  =  a.     And  as  we  have  taken  a  general  ex- 

Q 

pression,  we   conclude,   in  general,   that  a  vanishing  fraction  is  one 

which  assumes  the  form  of  g,  in  consequence  of  the  existence  of  a 

common   factor,  which  has  become  zero  by  a  particular  hypothesis, 

and  that  the  true  value  of  the  fraction  may  be  either  finite,  infinite,  or 

zero. 

227.  When  the  common  factor  is  apparent,  we  have  only  to  strike  it 
out  before  making  our  hypothesis,  and  we  get  at  once  the  true  value 
of  the  fraction.  But  there  are  many  expressions,  in  which  it  is  diffi- 
cult to  detect  the  common  factor,  and  it  then  becomes  necessary  to 
know  a  process  by  which  the  common  factor  may  be  discovered.  We 
will  illustrate  the  process  by  a  simple  example,  in  which  the  common 

X^ ft2  0 

factor  is  apparent.     Take  the  expression, =:  —  when   x  =  a. 

X  —  a         0 

Here,  the  assumed  value  of  x  is  a.     If,  however,  we  make  x  =  a  -{■  h, 

and  substitute  this  value  for  x  in  both  terms  of  the  fraction,  reduce  the 

result  to  its  lowest  form,  and  then  make  h  =  0,  it  is  evident  that  we 

will  have  done  nothing  more  than  attribute  to  x  the  value  a.     Making 

,1         ,    ,.,  ,.  ,  «'  +  2ah  +  li'  —  a"       2ali  +  A'       „ 

the  substitution,  we  get  ; = r i=z2a  -{-Ji^ 

2 

-  is  then  2a, 
when  X  =za. 

As  the  same  process  is  plainly  applicable  to  all  fractions  in  which 
the  terms  are  affected  with  iiumerical  exponents,  we  derive  for  such 
fractions  the  general 

RULE. 

Adrihufe  to  (hat  term  vpon  which  the  hypothesis  is  made  the  value 
V)hich  reduces  the  fraction  to  the  form  of  ^  plus  an  increment  h,  reduce 
the  result  to  its  lowest  form  and  then  make  h  =  0. 


VANISHING    TRACTIONS.  161 

In  the  above  example,  x  is  tlie  term  upon  which  the  hypothesis  is 
made,  and  a  the  value  which  reduces  the  fraction  to  the  form  of  g 
Hence,  by  the  rule,  x  =:a  +  h. 


228.  — EXAMPLES. 

1.  Find  the  value  of ■ =  —  when  x  =  a. 

X  —  a  0 

Ans.  3al 

2.  Iind  the  value  of  —. =  —  when  x  =  1. 

x'  —  X  —  ax  -\-  a        0 

Ans.    -. 

a  —  1 

o    I?-    wi,        1        pX^  -\-hx  +  X  +  h        0 

6.  lind  the  value  oi  —. =  —  when  x  =  —  1. 

x'  —  ax  -\-  X  —  a        0 

Ans. 


1  +a 

4.  Find  the  value  of  '^  ^  '.  '""  ' ' '^^^  =  ^  when  x  z=  m. 

Ans. 


X-  +  nx'  —  mx  —  nm-        f) 
x^  -\-  ax  —  mx  —  am  0 

m  (1  +  2«) 


a  -\-  111 

;;    r-   ^  .1         i        .  x^  —  \x^  —  xif  ■\- h/^        0      . 

5.  Find  the  value  of ■    ^ ^ ^  =  —  when  x  =  y. 


6.  Find  the  value  of  ^ -. —^  :=  —  when  x=.\. 

x^  —  x^  —  X  +  1         0 

Ans.    CO. 

7.  Find  the  value  of  the  fraction  — ^ '- —  =:  —  when 

x'  —  ox  —  ax  -\-  ab  0 

X  =  h.  Ans.  X  =  b  -{■  a. 

8.  Find  the  value  of  the  fraction  '— '—— =  —  when 

X  —  bx  —  ax  -{-  ab  0 

x=za.  Ans.  2a. 

14*  L 


162  YANISIIlNCi     FRACTIONS. 

The  two  last  results  arise  from  the  fact  of  the  given  fraction  being 
a  double  vanishing   fraction.       It   can   be    put   under   the   form    of 

^ ,  ^  ^' ,  ,  which  is  a  vanishing  fraction,  when  either  o:  =  h, 

(x  —  b)  (x  —  a) 

or  a;  =  n. 

229.  It  is  obvious,  in  all  these  examples,  that  x,  minus  the  value  of 
X,  which  makes  the  fraction  assume  the  form  of  ^,  is  the  common  fac- 
tor. If  we,  then,  divide  both  terms  of  the  fraction  by  this  common 
factor,  and  then  attribute  the  appropriate  value  to  x,  we  will  have  the 
true  value  of  the  vanishing  fraction.  But  there  are  many  expressions 
for  which  this  rule  fails,  as  will  be  seen  more  fully  hereafter. 

Since  it  is  often  difficult,  if  not  impossible,  without  the  aid  of  the 
differential  calculus,  to  ascertain  the  existence  of  a  common  factor,  if 
there  be  one,  it  becomes  important  to  have  a  simple  test  by  which  we 
can  tell  whether  £  indicates  indctcrmination  or  a  vanishing  fraction. 

Take  the  fraction  tt— ^^ ^,  which  becomes  [{  by  the  single  hypo- 

Q(.c  —  a)"' 

thesis  x  =  a.    It  is  evident  that  the  expression  is  a  vanishing  fraction, 

and  that  the  common  factor  is  some  power  of  x  —  a.     But  take   the 

'P  (x a)"' 

fraction  —j r^-,  which  becomes  Q  by  the  double  hypothesis  x  =  a, 

vj  (X  Uj 

and  x  =  h.  It  is  plain  that  the  expression  cannot  be  a  single  vanish- 
ing fraction  like  those  exhibited  in  the  first  six  examples ;  and  if  it  is 
a  double  vanishing  fraction  of  the  form  exhibited  in  Examples  7  and  8, 
it  cannot  be  a  true  solution  to  a  problem  of  the  first  degree,  since  x 
cannot  have  two  values.  We  then  conclude,  that  when  g  arises  from  a 
single  hypothesis  upon  a  solution  of  an  equation  of  the  first  degree,  it 
indicates  the  existence  of  a  common  factor.  But,  if  it  arises  from 
two  suppositions,  it  indicates  indetermination.  In  conformity  to  this 
rule,  we  have  interpreted  §,  in  the  problem  of  the  couriers,  to  indicate 
indetermination,  because  the  symbol  proceeded  from  the  double  hypo- 
thesis, VI  =:  0,  and  a  :=b. 


POWERS    AND    EXTRACTION    OF    ROOTS.  163 

FORMATION  OF  THE  POWERS  AND 
EXTRACTION  OF  ROOTS. 

230.  The  power  of  a  quantity  is  the  result  obtained  by  multiplying 
it  by  itself  any  number  of  times. 

Any  quantity  is  the  first  power  of  itself. 

If  a  quantity  be  multiplied  by  itself  once,  or  enters  twice  as  a  factor 
iu  the  result,  the  result  is  called  the  second  power  of  the  quantity. 
Thus,  2.2=  2!^,  or  4,  and  a  .  a  =a^,  are  the  second  powers  of  2 
and  a. 

If  a  quantity  be  multii^lied  by  itself  twice,  three  times,  four  times, 
&c.,  or  enters  into  the  result  as  a  factor  three  times,  four  times,  &c., 
the  result  is  called  the  third  power,  the  fourth  power,  &c.,  of  the  quan- 
tity. In  general,  the  number  of  multiplications  of  the  quantity  by 
itself  is  one  less  than  tht^quantity  which  designates  the  power,  and  the 
number  of  times  that  the  quantity  enters  as  a  factor  in  the  result,  is 
precisely  equal  to  that  quantity. 

The  quantity  which  designates  the  power  is  called  the  exponent  of 
the  power,  and  is  written  a  little  above  and  to  the  right  of  the  given 
quantity. 

Thus,  2*  =  4  is  the  second  power  or  square  of  2. 
2'  =  8  is  the  third  power  or  cube  of  2. 
2*  =  16  is  the  fourth  power  of  2. 
ay  is  the  ?/  power  of  a, 

and  indicates  that  a  has  been  multiplied  by  itself  y  —  1  times,  or  that 
a  enters  as  a  factor  y  times  in  the  expression  o>. 

When  no  exponent  is  written,  the  first  power  is  always  understood. 
Thus,  2  =  2',  and  a  +  i  =  (a  +  i)'. 

The  quantity  to  be  raised  to  a  power  may  be  expressed  numerically, 
or  by  letters,  and  may  be  entire  or  fractional,  positive  or  negative. 
And,  since  the  power  in  every  case  is  a  product,  we  may  define  the 
formation  of  a  power,  to  consist  in  finding  the  product  arising  from 
multiplying  the  quantity,  by  itself,  a  number  of  times  one  less  then 
that  indicated  by  the  exponent  of  the  power. 

The  poxcer  differs  from  an  ordinart/  prodxtcf,  then,  in  this  essential 
■particular,  all  the  factors  of  the  power  are  equal. 


1G4  FORMATION     OF     THE     POWERS     AND 

231.  The  root  of  a  quantity  is  tliat  quantity  which,  multiplied  by 
itself  a  certain  number  of  times,  will  produce  the  given  quantity. 

When  a  quantity,  multiplied  by  itself  once,  or  taken  as  a  factor 
twice,  gives  the  given  quantity;  it  is  called  the  square  root  of  the 
given  quantity.  Thus,  2  is  the  square  root  of  4,  because  2.2  =  4; 
and  a  is  the  square  root  of  a^,  because  a  .  a  ^=  a^. 

Raising  quantities  to  powers  is  called  Involution. 

Extracting  the  roots  of  quantities  is  called  Evolution. 

Involution  deals  in  equal  factors.  Evolution  finds  one  of  those  equal 
factors.  t 

232.  Involution  is  a  simple  process.  Evolution  is  more  difiicult, 
and  requires  particular  explanation.  "We  will  begin  with  the  simplest 
form  of  evolution,  the  extraction  of  the  square  root  of  whole  numbers, 
which  is  nothing  more  than  evolving  one  of  the  equal  factors  out  of  the 
product  of  two  equal  factors. 

It  is  evident  that  evolution  is  the  reverse  of  involution,  and  that  we 
cannot  extract  any  root  without  knowing  how  the  powers  of  that  root 
are  formed.  To  demonstrate  the  rule,  then,  for  the  extraction  of  the 
square  root  of  whole  numbers,  we  must  first  examine  and  see  how  the 
square  power  of  whole  numbers  is  formed. 

233.  The  first  ten  numbers  are 

1,  2,  3,    4,     5,     6,     7,     8,     9,     10. 
And  their  squares  are 

1,  4,  9,  16,  25,  36,  49,  64,  81,  100. 

Reciprocally,  the  numbers  of  the  first  line  are  the  square  roots  of  the 
corresponding  numbers  in  the  second  line.  We  see,  also,  that  the 
square  of  any  number  below  10  is  expressed  by  not  more  than  two 
figures.  That  is,  the  square  of  units  cannot  give  a  higher  denomina- 
tion than  tens.  So,  likewise,  it  may  be  shown  that  the  square  of  tens 
cannot  give  a  higher  denomination  than  thousands,  since  the  square  of 
99  is  9801. 

The  numbers,  1,  4,  9,  16,  &c.,  and  all  the  other  numbers  produced 
by  the  multiplication  of  a  number  by  itself,  are  called  pcr/t'c^  squares. 

We  see  that  there  are  but  nine  perfect  squares  between  1  and  100. 
The  square  roots  of  all  numbers  lying  between  1  and  100  will  be  found 
between  the  consecutive  roots  of  two  perfect  squares.  Thus,  the  square 
root  of  20  lies  between  the  consecutive  roots  4  and  5,  being  greater 
than  the  former,  and  less  than  the  latter.  The  square  root  of  26  lies 
between  the  consecutive  roots  5  and  6. 


EXTRACTION     OF    ROOTS.  165 

The  square  root  of  all  numbers  below  10,000  may  be  regarded  as 
made  up  of  tens  and  units.  Thus,  99,  the  square  root  of  9801,  is 
made  up  of  9  tens  and  9  units.  The  number  32  is  made  up  of  3  tens 
and  2  units.  We  have  seen  that  the  square  of  a  number  containing 
two  figures  could  not  give  a  higher  denomination  than  thousands ;  con- 
versely, the  square  root  of  thousands  cannot  give  a  number  containing 
more  than  two  figures ;  that  is,  a  number  containing  tens  and  units. 

284.  If,  then,  we  have  to  extract  the  square  root  of  a  number  con- 
taining more  than  three  figures,  and  less  than  five,  we  know  that  its 
root  must  contain  two  figures,  and,  therefore,  be  made  up  of  tens  and 
units.  Before  we  can  deduce  a  rule  for  the  extraction  of  the  root  of 
thousands,  we  must  know  how  thousands  are  derived  from  squaring 
tens  and  units. 

Let  a  represent  the  tens,  and  b  the  units,  which  enter  into  the  square 
root  of  thousands.  Then,  (a  +  by  =  a^  +  2ah  +  V^.  Hence,  thou- 
sands arc  made  up  of  the  square  of  the  tens  that  enter  into  its  root, 
plus  the  square  of  the  units,  plus  the  double  product  of  the  tens  by  the 
units. 

Let  us  square  the  number  34,  made  up  of  3  tens  and  4  units,  =  30 
4-  4.  Then,  30  corresponds  to  a  in  the  formula,  and  h  correspond.s 
to  4. 

Hence,  a^  =  (30^)      =     900, 

2rti  =  2  .  30  .  4  =  240, 
b'    =    (4f     =        16, 

Then,  (34)^  =  a'  -f  2ab  +  U'       =       1156. 

235.  If  we  were  required  to  extract  the  square  root  of  this  number, 
we  would  have  to  revei*se  the  process,  and  first  take  out  a*,  then,  2ah 
and  Z)l 

The  number,  1150,  belonging  to  the  denomination  of  thousands,  its 
root  must  contain  tons  and  units,  and 
we  must  first  get  out  the  tens  by 
extracting  the  square  root   of  the 
square  of  the  tens ;  that  is,  the  s/  «^- 
Now,  the  square  of  tens  will  give     2a  -\-  b 
at  least  three  figures,  therefore  the     60  -f-  4 
root  of  the  tens  cannot  be  sought  in 
the  two  right  hand  figures,  and  we, 
therefore,   separate  them  from  the 
others  by  a  dot,  and  the  given  number  is  then  separated  into  what  are 


a^  +  2ab  -f  // 

11 

.56 

a 

b 

9 

00 

=  a'     30  -f 

4 

T 

56 

=  2ab  -f  b' 

2 

40 
16 

0 

00 

Remaifider. 

16G  FOR-MATION     OF     THE     TOAVERS     AND 

called  pcrtoih  of  two  figures  each.  The  square  of  30  is  900,  and  the 
square  of  40  is  1600.  The  number,  115G,  falling  between  900  and 
1600,  its  root  must  contain  3  tens  plus  a  certain  number  of  units.  We 
then  extract  the  greatest  square  contained  in  the  left  hand  period,  1100, 
and  set  the  root,  30,  on  the  right,  after  the  manner  of  a  quotient  in 
division.  We  have  now  found  a,  square  it,  and  subtract  a^  or  900, 
from  1156,  and  the  remainder  must  be  2oi  +  1'^:  in  the  present  instance 
equal  to  256.  The  remainder,  2ah  +  1j\  can  be  written  (2a  +  l)h; 
and  it  is  evident  that,  to  find  h  accurately,  we  must  divide  by  (2a  +  h). 
But,  as  the  term  h  within  the  parenthesis  is  unknown,  we  are  com- 
pelled to  use  2a  as  the  aj)proximate  divisor.  We  write  2a,  or  60,  on 
the  left  as  a  divisor,  and  divide  256  by  it,  and  set  the  quotient  h,  or  4, 
on  the  right  of  the  root  found,  and  also  on  the  right  of  the  divisor. 
Now,  multiply  the  two  terms  on  the  left,  2a  -f  J,  or  60  +  4  by  Z»  or  4, 
and  we  evidently  form  2a^  +  ?/,  the  parts  entering  into  the  remainder, 
256.  Subtracting  the  two  products  thus  formed  from  256,  we  find  no 
remainder.     Hence,  34  is  the  exact  root  of  1156. 

We  have  separated  the  tens  from 
the  units,  to  let  the  beginner  see  that 
he  is  really  taking  a^  -}-  2ah  +  1/  in 
succession,  out  of  the  given  number, 
2a  +  h  \  2  56  =  2a6  +  V^  1156.  But  we  might  have  indicated 
6  4  I  2  56  =  2ah  +  V  their  separation  by  a  point  above,  and 
written  3  4,  instead  of  30  +  4,  and 
6  4  instead  of  60  +  4.  When  the  beginner  is  familiar  with  the  princi- 
ples, he  may  omit  the  dots,  -which  are  intended  to  guard  him  against 
confounding  the  tens  with  the  units.  He  must  observe,  however,  that 
this  divisor  being  the  double  product  of  tens,  is,  itself,  tens,  and,  there- 
fore, if  written  out  in  full,  would  contain  a  cj'pher  on  its  right.  And 
since,  in  dividing  by  a  number  whose  right  hand  figure  is  0,  we  point 
ofi"  that  figure  from  the  right  of  the  divisor,  and  also  point  oif  the  right 
hand  figure  of  the  dividend,  we  must  be  careful  to  do  this  in  dividing 
by  the  double  product  of  the  tens.  In  the  present  instance,  the  right ' 
hand  figure,  6,  of  the  remainder  256,  must  be  separated  from  the  other 
two  figures,  since,  in  using  6  as  the  divisor  instead  of  60,  we  have,  in 
fact,  pointed  off  0  from  the  right  of  the  divisor. 

236.  -There  is  one  point  of  considerable  importance  that  needs  some 
examination.  In  getting  the  second  figure  of  the  root,  we  used  2a  as 
tlie   approximate    divisor   of  the   remainder,  2al>  -\-  1/  =z  (2a  +  li)  h, 


u'-\ 

:-  2ah 

-Vh' 

11 

.  56 

a  -\-h 

9 

00  = 

=  a^ 

8  ■  4 

2 

"56"; 

=  2a6 

+  1' 

2 

56  = 

=  2ah  +  V 

EXTRACTION    OF    ROOTS.  167 

whereas,  the  true  divisor,  to  find  h,  is  plaiuly  2a  +  h.  Our  divisor 
being  too  small,  the  quotient,  which  is  the  second  figure  of  the  root, 
can  never  be  too  small,  but  may  be  too  great.  It  is  plain  that,  when  h, 
in  the  expression  (2a  +  h),  is  very  small  in  comparison  with  '2a,  it 
may  be  neglected.  But  b  may  be  so  large  that  the  omission  of  it 
will  give  too  great  a  quotient.  The  square  of  35  is  1225 ;  the  square 
root  of  1225  is  then,  of  course,  35.  Now,  if  we  proceed  to  extract 
the  square  root  of  1225,  the  remainder,  after  taking  out  a",  or  900, 
will  be  found  to  be  325  =  2ah  -\- V  =  300  +  25.  And  we  sec  that 
the  square  of  the  units  has  added  2  tens  to  'lah,  the  double  product  of 
the  tens  by  the  units.  When,  therefore,  we  point  off  5  from  the  right 
of  32  5,  and  divide  32  by  6,  it  is  plain  that  the  dividend,  which  ought 
to  be  2ah  (if  the  divisor  is  2a),  is  too  great  by  2  tens.  The  quo- 
tient, then,  would  be  too  great,  if  the  2  tens  added  were  divisible  by 
the  divisor.  Then  the  second  figure  of  the  root  would  be  augmented 
improperly  by  the  quotient,  arising  from  dividing  the  2  tens  impro- 
perly added  by  G,  or  2a.  In  the  present  case,  however,  if  we  omit  the 
2  tens,  and  divide  30  by  G,  we  get  the  same  quotient,  5,  as  when  the  2 
tens  are  retained;  their  addition  has    not,   then,  affected  the  result. 

But,  square  19,  and  we  get  3G1. 
In  this  case,  h,  in  the  expression 
(2a  -f  6),  is  not  small  in  comparison 
with  2a,  and  cannot,  therefore,  be 
neglected  without  affecting  the  re- 
sult. Now,  if  we  use  2a  as  the 
divisor  of  the  remainder,  2ai  -f  W, 
to  find  h,  we  ought  to  use  lah  alone 
as  the  dividend;  and,  therefore,  if 

we  use  the  whole  of  the  expression,  2ah  -\-  V,  our  dividend  is  too  great. 
Dividing  26  by  2,  the  quotient  is  13,  which  is  plainly  absurd  for  the 
units  of  the  root.  But,  we  see  that  180,  or  lah,  ought  to  have  been 
the  dividend  corresponding  to  20,  or  2a,  as  a  divisor.  The  26  tens  is 
then  too  great  by  8  tens ;  and,  since  the  8  tens  added,  give  4  for  a 
quotient  when  divided  by  the  2  tens  of  the  divisor;  the  second  figure 
of  the  root  is  too  great  by  4,  and  we  must  write  9  as  that  figure,  and 
not  13.  The  8  additional  tens  in  the  2G  tens  come  from  the  square  of 
the  units,  and  being  divisible  by  the  divisor,  2a,  have  improperly  aug- 
mented the  second  figure  of  the  root.  Had  the  8  tens  not  been  divisible 
by  2rt,  the  second  figure  of  the  root  would  not  have  been  increased  at 
all,  and  the  quotient  of  lah  -\-  W  by  2a  would  have  truly  been  the 


"•-  +  : 

lah  +  />2 

3G1 

a  -{■  b 

100: 

=  a^ 

19 

+  h 

261: 

=  2ah  + 

b' 

2  9 

18  0: 

=  2ah 

81: 

=  1'' 

00  0 

Remainder. 

168  FORMATION    OP    THE    POWERS    AND 

second  figure  of  the  root.  In  general,  whenever  the  square  of  the  units 
(i^)  incorporate  into  the  remainder,  2ab  +  Z;'^,  tens  which  are  exactly 
divisible  by  2a,  the  divisor,  the  second  figure  of  the  root  will  be  too 
great,  and  must  be  diminished  by  the  quotient  of  the  incorporated  tens 
by  the  2a  of  the  divisor. 

237.  The  foregoing  course  of  reasoning  has  shown  that  the  second 
figure  of  the  root  may  be  too  great,  and  the  cause  of  its  being  too  great; 
but,  since  the  units  of  the  root  are  unknown,  the  number  of  the  tens 
proceeding  from  their  square,  that  are  incorporated  with  2ah,  cannot 
be  known.  We  must,  then,  in  practice,  form  the  product  of  2a  +  5 
on  the  left  by  the  second  figure,  I,  of  the  root,  and  compare  the  result 
with  the  remainder.  If  the  product  is  greater  than  the  remainder,  the 
second  figure  must  be  diminished  until  the  product  is  equal  to  the  re- 
mainder, or  smaller  than  it.  If  the  given  number  is  an  exact  square, 
its  root  will  be  exact,  and  the  product  will  be  exactly  equal  to  the  re- 
mainder. When  the  root  is  not  exact,  the  product  must  be  made  less 
than  the  remainder. 

The  preceding  principles  enable  us  to  deduce  for  the  extraction  of 
the  square  root  of  whole  numbers,  embraced  between  100  and  10,000, 
the  following 

RULE. 

I.  Separate  the  two  rigid  hand  figures  from  the  other  figxires  or 
figure  of  the  given  numher,  and  find  the  greatest  square  contained  in 
the  left  hand  period,  which  may  contain  hut  one  figure. 

II.  Set  the  root  of  this  greatest  sq%iare  on  the  right,  after  the  manner 
of  a  quotient  in  division.  Sichtract  the  square  of  the  root  thus  found 
from  the  first  period,  and  annex  the  second  period  to  the  remainder. 

III.  Double  the  root  found  and  place  it  on  the  left  for  a  divisor. 
Seek  how  often  the  divisor  is  contained  in  the  remainder,  exclusive  of 
the  right  hand  fig%ire,  and  place  the  quotient  on  the  right  of  the  root 
already  found,  separated  from  it  hy  a  dot  above.  Place  it  also  on  the 
right  of  the  divisor,  separated  from  it  in  like  manner. 

IV.  Midtiply  the  divisor  thus  augmented  by  the  second  figure  of  the 
root,  and  suhtract  the  product  from  the  first  remainder.  If  there  is  no 
remainder,  the  root  is  exact.  Jf  the  product  exceed  the  first  remainder, 
the  second  figure  of  the  root  must  be  diminislied  until  the  product  is 
equal  to  or  smaller  than  the  first  remainder. 


EXTRACTIOX    OF    ROOTS. 


EXAMPLES. 


1.  Extract  the  square  root  of  225.  Ayis.  15. 

2.  Extract  the  square  root  of  7569.  Ans.  87. 

3.  Extract  the  square  root  of  2025.  Ans.  45. 

4.  Extract  the  square  root  of  841.  Ans.  29. 

5.  Extract  the  square  root  of  2500.  Ajis.  50. 

6.  Extract  the  square  root  of  7921.  Ans.  89. 

7.  Extract  the  square  root  ot  9801.  Atis.  99. 

8.  Extract  the  square  root  of  4096.  Ans.  64. 

9.  Extract  the  square  root  of  5476.  Ans.  74 
10.  Extract  the  square  root  of  7056.  Ans.-  84. 

238.  We  have  seen  that  the  second  figure  of  the  root  has  frequently 
to  be  diminished.  We  may  diminish  it  too  much,  and  it  becomes  ne- 
cessary to  know  when  we  have  made  the  second  figure  too  small.  The 
test  of  this  depends  upon  the  principle,  that  the  difference  between  two 
consecutive  squares  is  equal  to  twice  the  smaller  number  plus  unity. 

Let  a  =.  smaller  number. 

Then,  a  +  1  =  consecutive  number,  or  the  number  just  above  a. 

And  (a  +  1)^  =  a^  +  2a  +  1. 

Their  difference  is  2a  -j-  1,  as  enunciated. 

Now,  when  there  is  a  remainder,  after  finding  the  second  figure  of 
the  root,  and  subtracting  the  product  of  it  by  the  quantity  on  the  left 
from  the  first  remainder,  it  is  evident  tbat  the  second  remainder  ex- 
presses the  excess  of  the  given  number,  which  we  may  regard  as 
(a  4-  l)'^,  over  the  square  of  the  two  figures  found.  If,  then,  the 
second  remainder  be  exactly  twice  the  root  found  plus  unity,  it  is  evi- 
dent that  the  root  found  is  o,  and  that  the  second  figure  of  the  root  can 
be  increased  by  unity. 

To  illustrate,  suppose  6  to  be  the  square  root  of  49,  (a  -{-  1)^ 

then  the  remainder  being  equal  to  twice  the  root  found      49  I  a 

plus  unity,  the  root  can  be  increased  by  unity.     In      36  =  a^  I  6 
general,  whenever  the  remainder  exceeds  twice  the      1 ."  —  2a  -j-  1 
root  found  plus  unity,  the  root  can  be  augmented  by 
unity.     If  the  remainder  is  exactly  equal  to  twice  the  root  found  plus 
15 


170  FORMATION     OP     THE     POWKRS     AND 

unity,  the  root,  increased  by  unity,  will  be  the  exact  root  of  the  given 
number. 

239.  This  rule  is  of  importance  in  finding  the  square  root  of  imper- 

fect squares.    Let  it  be  required  to  find  the  square  root 

1  5G  1 12     of  156.     We  find  a  remainder  12,  and  a  root  12.     Is 

1  12  the  greatest  root  contained  in  156  ?     Is  the  root  12 

22  I     56  plus  a  remainder,  or  13   plus  a  remainder?     By  the 

44  principle  just  demonstrated,  the  true  root  of  156  must 

12  be  12  plus  a  remainder,  because  the  second  remainder 

is  not  double  the  whole  root  found,  plus  unity. 

240.  This  principle  also  enables  us  to  pass  from  the  square  of  a 
number  to  the  square  of  a  consecutive  number  without  raising  the 
second  number  to  the  square  power.  We  have  only  to  represent  the 
smaller  number  by  a,  then  the  consecutive  number  will  be  a  +  1,  and 
its  square  must  exceed  a?  by  2a  +  1. 

Thus,  (100)=^  =r  a^  =  10000. 

a^         (2a  +  1) 
Then,  (101)^  ==  (a  +  1)'  =  10000  -}-  200  +  1  ==  10201. 

(a  +  1)^  a^  (2a  +  1) 

So,  also,  (102)-^  =  (101  +  If  =  10201  +  202  +  1  =  10404. 

The  following  are  incommensurable  numbers. 


1.  Find  the  square  root  of  1720.  J.ns..  41  + 

2.  Find  the  square  root  of  1445.  Ans.  38  + 

3.  Find  the  square  root  of  6411.  Ans.  80  + 

4.  Find  the  square  root  of  5555.  Ans.  74  + 

5.  Find  the  square  root  of  1755.  Ans.  41  + 

6.  Find  the  square  root  of  1960.  Ans.  44  + 

7.  Find  the  square  root  of  7777.  Ans.  88  + 

8.  Find  the  square  root  of  6666.  Ans.  81  + 

241.  If  we  square  any  number,  as  12  and  55,  containing  two  figures, 
and  made  up  then  of  tens  and  units,  the  square  will  contain  two  periods, 
counting  from  the  right,  and  it  is  plain  that  the  tens  can  only  be 
sought  in  the  periods  on  the  left.     If  we  square  a  number  made  up  of 


EXTRACTION     OF    ROOTS.  171 

hundreds,  tens,  and  units,  the  square  'will  contain  three  periods,  and 
the  hundreds  can  only  be  found  in  the  left  hand  period,  and  the  tens 
only  in  the  second  period,  annexed  to  what  is  left  of  the  first  periods 
after  the  square  of  the  hundreds  has  been  taken  from  it.  In  general, 
the  number  of  periods  in  the  given  number  always  indicate  the  number 
of  figure  places  in  the  root,  and  each  figure  of  the  root  has  its  appro- 
priate period  or  periods. 

The  principles  that  have  been  demonstrated  for  the  extraction  of  the 
square  root  of  numbers  between  100  and  10,000  can  readily  be  ex- 
tended to  any  numbers  whatever.  Let  a  represent  the  highest  denomi- 
nation in  the  root,  and  s  all  the  succeeding  denominations  in  the  root. 
Then  the  number  itself  will  be  expressed  by  (a  -f  s)-  =  o?  -\-  2as  -\-  s^, 
an  analogous  expression  to  («  +  by  =  a-  -j-  2ah  +  I/,  and  differing 
only  in  its  more  general  significance.  The  a  in  one  formula  is  not  re- 
stricted to  represent  tens,  as  it  is  in  the  other,  but  may  represent  hun- 
dreds, thousands,  millions,  &c.  j  and  the  s  of  the  first  formula  is  not 
restricted  to  represent  units  only,  but  may  represent  tens  and  units, 
hundreds,  tens,  and  units,  &c. 

Let  us  square  155  by  means  of  the  formula. 

Then,  a  =  100,  and  s  =  55. 

a'  =  (100)-  =  10000, 
2as  =  200'55  =11000, 
s'    =(55/      =    3025, 

Hence,  (155)^         =         24025 

We  see  from  the  formula  that  a,  it  matters  not  what  may  be  its  de- 
nomination, must  first  be  found;  and,  that  after  its  square  has  been 
subtracted  from  the  given  number,  the  remaiiuler  will  be  2t(H  -f  s"  = 
(2a  +  s)s. 

The  number,  24025,  being  greater  than  lOOOO,  its  root  will  be 
greater  than  100,  and,  therefore,  a- 

canuot   be  found  in  the  two  right  «"  +  2('s  +  s^ 

hand  periods.     We  seek  it  in  the  2  40  25         |  a  +  s 

period  on  the  left,  and  after  placing  1  00  00  I    1  55 

it  on  the  right,  and  subtracting  its     2a  -f  s  I  1  40  25  =  2as  +  0^ 
square  from  the  given  number,  have        255    I  1 11  00  =  2as 
14025  for  a  remainder.     We  cut  off  30  25  =  s' 

the  right  hand  figures,  because  2,  the  0  00  00     Remainder, 

approximate   divisor,  is  really  200; 
and,  after  trial,  we  find  55  to  be  the  right  hand  figures  of  the  root. 


172  FORMATION     OF     THE     POWERS     AND 

The  55  is  set  on  the  right  of  the  root  ah-eady  found,  and  also  on  the 
right  of  the  divisor.  The  product  of  255  by  55  is  14025,  and  there 
is,  consequently,  no  second  remainder,  and  the  root  is  exact. 

2-42.  The  approximate  divisor  is  always  large  for  numbers  above 
10000,  and  s  can  only  be  found  by  repeated  trials.  But  the  above 
process  can  be  greatly  simplified  by  observing  that,  since  2a  enters 
into  s,  representing  several  denominations,  it  must  enter  into  each  de- 
nomination separately.  Thus,  in  the  foregoing  example,  2a,  or  2, 
being  a  multiplier  of  55,  is  a  multiplier  of  the  first  5,  regarded  as  5 
tens,  or  50,  and  of  the  second  5,  regarded  as  units.  We  might,  then, 
have  found  the  5  tens  and  the  5  units,  separately  taking  care  to  write 
the  one  after  the  other,  so  as  to  make  their  denomination  distinct.  In 
the  present  example,  2as  being  equal  to  2  .  55,  is,  of  course,  equal  to 
2  (^50  +  5)  J  s  may  be  regarded  as  a  single  term,  55,  to  be  found  at 
once,  or  it  may  be  regarded  as  made  up  of  50  and  5,  separate  terms,  to 
be  found  separately.  But,  if  the  second  and  third  figure  of  the  root  be 
found  separately  as  independent  numbers,  they  must  be  sought  for  in 
their  appropriate  periods.  It  will  simplify  the  work  when  we  proceed 
in  this  manner  to  subtract  the  square  of  a  from  the  left  hand  period, 
and  bring  down  each  term  in  succession.  In  this  case,  since  we  make 
two  terms  of  s,  let  s  ^  s'  -f  s" .  Then  the  root  will  be  a  +  s"  -f  s", 
and  the  number  will  be  (a  +  s"  +  s")^,  which,  by  performing  the  mul- 
tiplication indicated,  will  give  us  a^  +  2os'  -f-  s'^  -f-  2as"  +  2sV'  +  s"^ 
=  a"  +  2as'  +  s'^  +  2  {a  +  s')  s"  +  i>"\     When  we  subtract  o^  from 

the  left  hand  period, 
we  have,  in  fact, 
subtracted  10000 
from  the  given  num- 
ber. After  the  se- 
cond period  has 
been  annexed  to 
the  remainder,  the 
140  truly  represents 
14000,  and,  since 
the  zero,  on  the  right  of  140,  belongs  to  the  denomination  of  hundreds, 
it  must  be  separated  from  the  14  when  we  come  to  seek  for  the  tens  of 
the  root,  because  /,  the  tens,  is  sought  for  in  2as'  by  using  2a  as  the 
approximate  divisor.  Now,  2as'  must  be  at  least  thousands,  and, 
therefore,  the  denomination  of  hundreds  does  not  contain  s'.  We  write 
s,  when  found,  on  the  right  of  the  first  term  of  the  root  and  also  on 


2a  +  s' 
2  5 

(«  +  s'  +  s"f 
240'25              a  -f «'  -f  s" 
1    .    .    =a'      l'5'5 
1  40      —  2as'  -h  s'2  ^  &c. 
12  5      =  2as'  +  s'2 

2  (a +  0+5" 
30  5 

1  52  5  =  2  (a  +  s')  s"  + 
1  52  5  =  2  (a  +  s')  .s"  + 
0         Remainder. 

EXTRACTION     OF    ROOTS.  173 

the  right  of  the  divisor.  Multiplying  the  divisor  thus  augmented  by 
s'  in  the  root,  and  subtracting  the  product  from  the  first  remainder,  we 
will  plainly  have  left,  after  annexing  the  next  period,  2  (a  -f  s').s"+  s"^. 
And  we  see  that  the  approximate  divisor  to  find  s"  is  2  (a  +  s').  The 
whole  root  already  found  mu.st  then  be  doubled  and  used  as  our  approxi- 
mate divisor.  The  right  hand  figure  of  1525  is  cut  off,  because 
2  (a  +  s')s"  gives  at  least  hundreds.  After  s"  is  found,  we  multiply 
'1  (a  -\-  s')  by  it,  and  have  no  remainder;  the  root  is  then  exact.  We 
have  used  the  broken  line,  in  the  above  example,  after  each  minuend 
and  subtrahend  to  indicate  that  there  were  other  numbers  to  follow 
them. 

(a  +  i/  +  a"  +  a'"/ 

2  98'59"84  i  a  +  s'  +  s"  +  «'" 


1  98  =  2as'  +  s"  +  &c. 

1  89  rr=  2r,s'  +   s" 


9 


2a  + 
27 
(„  +  /)  +  ,"    9  5  9      =2{a+  sy  +  s"  -f  &c. 


34  2 

2(a  +  s'  +  s")  +  i 
3418 


684      =  2  (a  +  s')s"  + 


2  7  58  4  =zz  2  (a  +  s'  -f  /')■•>■'"  +  «""  +  &c. 
2  7584==  2  {n  +  a'  +  6-").s"'  +  s"'\ 


If  wc  have  a  number  made  up  of  4  periods,  as  2  98  59  84,  we  know 
that  its  root  must  contain  4  figure  places,  which  may  be  represented  by 
a,  s',  ,s",  and  s'".  And  the  given  number  must  then  be  equal  to 
(a  4-  s'  +  s"  +  s'")^  =  a""  +  2as'  +  s"  +  2  (a  +  s')s"+  s"^  +  2  (a+sf 

We  sec,  then,  that  2a  is  the  approximate  divisor  to  find  .s',  the 
second  figure  of  the  root ;  (2a  +  s')  the  approxim-ate  divisor  to  find  s", 
the  third  figure  of  the  root;  and  (2(a-\-sf)  +  s")  the  approximate  di- 
visor to  find  s'",  the  units  of  the  roots.  In  other  words,  we  see  that 
the  whole  root  found  has  to  be  doubled  to  find  each  figure  of  the  root 
succeeding  those  already  found.  It  is  evident,  too,  that  the  right  hand 
figure  of  each  of  the  successive  remainders  must  be  cut  off  previous  to 
dividing  by  2a,  2  (a  +  s'),  &c.,  because  in  all  these  remainders  that 
figure  is  of  too  low  a  denomination  to  make  any  part  of  the  product  of 
the  figure  of  the  root  sought  by  the  approximate  divisor.  Thus, 
8,  which  belongs  to  the  denomination  of  tens  of  thousands,  cannot 
be  a  part  of  the  product  arising  from  multiplying  2a,  or  2000,  by  s', 
which  is  hundreds.  This  product  must  give  at  least  hundreds  of 
thousands. 
15* 


174  FORMATION     OF     THE     POWERS     AND 

243.  It  is  plain  tliat^  from  the  manner  in  which  tlie  square  of  any 
number  of  terms  is  formed,  the  foregoing  demonstrations  for  numbers 
having  two,  three,  and  four  periods  are  general ;  and  we,  therefore,  have 
for  the  extraction  of  the  square  root  of  any  number  whatever,  the 
general 


I.  SejxtJ'ate  the  given  number  into  periods  of  two  figures  each,  begin- 
ning on  the  right;  the  left  hand  period  may  contain  but  one  figure. 

II.  Find  the  greatest  square  contained  in  the  left  hand  period,  and 
set  its  root  on  the  right,  after  the  maimer  of  a  qxiotient  in  division. 
Subtract  the  square  of  the  root  found  from  the  left  hand  period,  and  to 
the  remainder  annex  the  second  period,  and  use  the  number  tints  found 
as  a  dividend. 

III.  Doidjle  the  root  found  and  p>^(-^ce  it  on  the  left  for  a  divisor. 
Seek  how  often  the  divisor  is  contained  in  the  dividend  exclusive  of 
the  right  hand  figure,  and  place  the  qtiotient  on  the  right  of  the  root 
already  found,  and  separate  the  two  figures  by  a  point.  Set  the  quo- 
tient also  on  the  right  of  the  divisor  and  separate  in  like  manner. 

IV.  Multiply  the  divisor  thus  increased  by  the  second  figure  of  the 
root,  subtract  the  product  from  the  dividend,  and  to  the  remainder 
annex  the  second  period  of  the  given  number.  Use  the  remainder  and 
annexed  period  as  a  new  dividend. 

V.  Double  the  whole  root  found  for  a  new  divisor,  and  continue  the 
operation  as  before  until  all  the  periods  are  brought  down. 

Remarks. 

244.  If  the  last  remainder  is  zero,  the  given  number  is  a  perfect 
square.  But,  if  the  remainder  is  not  equal  to  zero,  we  have  only  found 
the  entire  part  of  the  root  sought,  and  the  given  number  is  incommen- 
surable. 

245.  If  we  take  155,  the  square  root  of  24025,  we  observe  that  15 
has  been  derived  from  the  two  left  hand  periods.  We  might,  then, 
after  finding  5,  have  squared  15,  and  subtracted  its  square  from  240. 
So,  after  finding  the  last  5  of  the  root,  we  have  subtracted  the  square 
of  155  from  the  three  given  periods.  In  general,  when  we  have  found 
two  figures  of  the  root,  or  three  figures,  or  four  figures,  &c.,  we  may 
subtract  their  square  from  the  two  left  hand  periods,  or  the  three  left 
hand  periods,  &c. 


EXTRACTION     OF     ROOTS.  175 


GENERAL   EXAMPLES. 


1.  Extract  the  square  root  of  16008001.  Ans.  4001. 

2.  Extract  the  square  root  of  4937281.  Ans.  2222. 

3.  Extract  the  square  root  of  1111088889.  .4ns.  33333. 

4.  Extract  the  square  root  of  1975304G913G.  Ans.  444444. 

5.  Extract  the  square  root  of  36000024000004. 

Ans.  6000002. 

6.  Extract  the  square  root  of  12.59631362889.      Ans.  1122333. 

7.  Extract  the  square  root  of  15241383936.  Ans.  123456. 

8.  Extract  the  square  root  of  16080910030201. 

Ans.  4010101. 

9.  Extract  the  square  root  of  123456787654321. 

Ans.  11111111. 

10.  Extract  the  square  root  of  12345678987654321. 

Ans.   111111111. 

11.  Extract  the  square  root  of  308641358025.  Ans.  555555. 


OF   INCOMMENSURABLE   NUMBERS. 

246.  An  incommensurable  number  is  one  whose  indicated  root  can- 
not be  exactly  extracted.  Thus,  the  \/2,  V8,  and  \/27  are  incom- 
mensurable numbers.  Such  numbers  are  also  called  irrational  num- 
bers, ajid  sometimes  surds. 

We  have  indicated  (Art.  240,)  that  the  roots  of  imperfect  square 
powers  were  not  complete  by  writing  the  sign  +  after  the  entire  parts 
of  those  roots.  So  we  may  write  the  -v/  5  :=  2  -f .  The  number  5 
lying  between  4  and  9,  the  square  roots  of  which  are  2  and  3,  its  own 
root  will  be  greater  than  2,  and  less  than  8.  May  not  this  root,  then, 
be  expressed  by  some  fraction  whose  value  is  greater  than  2,  and  less 
than  3,  such  as  |,  or  |  ?  May  not  the  roots  of  all  imperfect  square 
powers  be  expressed  by  vulgar  fractions  in  exact  parts  of  unity  ? 

To  prove  that  this  cannot  be,  we  will  demonstrate  a  theorem  upon 
which  depends  the  proof  of  its  absurdity. 

Theorem. 

247.  Every  number^  P,  which  will  exactly  divide  the  product, 
A  X  B,  of  two  numbers,  and  which  is  prime  with  respect  to  one  of 
them,  will  divide  the  other. 


176  FORMATION     OF    THE    POWERS    AND 

Let  A  be  the  number  with  which  P  is  prime.     Let  Q  be  the  qua 
tieut  arising  from  dividing  AB  by  P,  then  -t— -  =  Q.     We  may  put 

this  equation  under  the  form  A  X  ^tj  =  Q.     Now,  Q  is,  by  hypothesis, 

an  entire  number;  the  second  member  being  a  whole  number,  the  first 
member  must  also  be  a  whole  number,  else  we  would  have  an  irreduci- 
ble fraction  equal  to  a  whole  number,  which  is  absurd.     The  product 

of  A  into  r—  has  then  to  be  entire ;  now,  A  itself,  is  entire  and  prime 

T) 

with  respect  to  P,  hence,  —  must  also  be  entire.     For  a  whole  number, 

multiplied  by  a  fraction,  can  only  give  an  entire  product  when  the 
whole  number  is  divisible  by  the  denominator  of  the  fractfon  into  which 
it  is  multiplied.  Thus  4,  a  whole  number,  multiplied  by  the  fraction 
#,  gives  an  entire  product,  because  4  is  divisible  by  2.  But  5  into  | 
does  not  give  an  entire  product,  because  5  is  not  divisible  by  2.  Now, 
the  whole  number  A  is,  by  hypothesis,  not  divisible  by  the  denomina- 

B  B 

tor,  P,  of  the  expression  p- ;  the  product  of  A  by  ^15-  cannot,  then,  pos- 
sibly be  equal  to  the  whole  number,  Q,  unless  B  is  divisible  by  P. 

248.  We  are  now  prepared  to  show  that  the  square  root  of  an  im- 
perfect square,  such  as  5,  cannot  be  expressed  by  a  fraction.     If  the 

root  of  5  can  be  expressed  by  a  fraction,  let  —  be  that  fraction,  a  being 

greater  than  h,  and  prime  with  respect  to  it.  We  assume  a  and  b  to 
be  prime  with  respect  to  each  other,  because,  otherwise,  their  quotient 
would  be  a  whole  number,  and  we  know  that  the  root  of  an  imperfect 
power    is    found    between    two  whole    numbers.       We   have,   then, 

y/5'=  — .  From  which,  by  squaring  both  members,  there  results 
5  =  -y^-  The  first  member  of  this  equation  being  a  whole  number, 
the  second  member  must  be  a  whole  number  also.     But  —^  cannot   be 

a  whole  number  unless  a^  is  divisible  by  I ;  for,  to  divide  a^  by  b^,  is 
to  divide  it  twice  by  b.  Of  course,  then,  if  a^  is  not  divisible  by  b,  it 
cannot  be  divisible  by  b^.  But  a^  is  not  divisible  by  h,  for,  by  the  fore- 
going theorem,  a  number  which  exactly  divides  the  product  of  two 
factors,  and  is  prime  with  respect  to  one  of  thenj,  must  divide  the 


EXTRACTION     OF    ROOTS.  177 

otter.  Now  tte  factors  of  o^  are  a  and  a ;  6  is  prime  with  respect 
to  the  first  factor,  it  must  then  divide  the  second  in  order  to  give  an 
entire  quotient.  This  is  plainly  impossible,  since  the  second  factor  is 
the  same  as  the  first;  a^  is,  then,  not  divisible  by  h;  still  less,  then, 

can  a^  be  divisible   by  Ir.     The  equation  5  =  —  must,  therefore,  be 

absurd ;  but  that  equation  has  been  truly  derived  from  the  equation 

v/?  =  -r-     A  correct  algebraic  operation  has  led  to  an  absurd  result : 
b 

but  this  can  only  be  so  when  the  assumption  at  the  outset,  upon  which 
the  operation  is  based,  is  absurd.  The  assumption,  •>/~o=  —  was  ab- 
surd, and  it  has  led  to  an  absurd  result. 

The  foregoing  reasoning  has  been  in  no  way  dependent  upon  the 
fact  that  5  was  the  particular  imperfect  power  under  consideration,  and 
is,  therefore,  general.  We  conclude,  then,  that  the  square  root  of  no 
imperfect  power  can  be  expressed  by  an  exact  fraction. 


EXTRACTION  OF  THE  SQUARE  ROOT  OF  FRACTIONS. 

249.  Since  the  square  of  a  fraction  is  formed  by  squaring  the  nume- 
rator and  denominator  separately,  it  follows  that  the  square  root  of  a 
fraction  must  be  taken  by  extracting  the  square  root  of  the  numerator 
and  denominator  separately. 

Thus,  ^^  =  'j,  since  -5X3=  |. 


,0,  also,  \/'jr,  =  j,  because  j  X  -  = -^, 


We  may  remark  that  all  square  roots  can  be  affected  with  either  the 
positive  or  negative  sign,  if  these  roots  are  regarded  as  algebraic  quan- 
tities.    Thus,  the  \/|  may  be  either  +  |,  or  — |,  because  — |  X  — 

„         .       „       ,  /  a^  1.1  (I  a        .  a 

I  =  |.     ho,  also,  \  /  71  iii'^y  be  either  -f  — ,  or  —  — ;  since  —  y, 

multiplied  by  itself,  gives  y^. 


1.  Extract  the  square  root  of  ^^. 

2.  Extract  the  square  root  of  /g.  Ans.  |,  or  —  |. 

3.  Extract  the  square  root  of  i.  Ans.  ^,  or  — |. 

M 


An^. 

^or-J. 

An?;. 

..or  — 1. 

Ans. 

4,  or  — #. 

Alls. 

|,or-i. 

Ans. 

ior-|. 

Ans.    J^- 

Sor-V- 

Ans. 

5,  or  — 5. 

ITS  FORMATION    OF    THE    TOW  Ell  S    AND 

4.  Extract  the  square  root  of  -1. 

5.  Extract  the  square  root  of  ^V- 

6.  Extract  the  square  root  of  ~-f. 

7.  Extract  the  square  root  of  ^-^. 

8.  Extract  the  square  root  of  f  1. 

9.  Extract  the  square  root  of  lY- 
10.  Extract  the  square  root  of  ■*^^^. 

250.  Those  examples  show  that  the  positive  square  roots  of  proper 
fractious  are  greater  than  the  fractions  themselves.  Thus,  ^  {-  :=  h. 
The  reason  of  this  is  plain :  the  numerator  of  a  proper  fraction  being 
less  than  the  denominator,  is  not  diminished  proportionally  so  much  as 
the  denominator  by  the  extraction  of  the  square  root.  Let  a^  and  1/  be 
two  unequal  squares,  and  let  a^  ^  b^;  then,  a  ^  h,  when  a  and  b  are 
both  positive.  Extracting  the  square  roots  of  a^  and  b'^,  we  have 
n  and  b  for  the  roots.  The  extraction  of  the  root  has  then  diminished 
fi^  a  fold,  and  ?/  only  b  fold.  The  greater  quantity  has,  then,  been 
diminished  the  most. 

251.  The  positive  roots  of  improper  fractions  aVe  less  than  the  frac- 
tions themselves,  because  the  extraction  of  the  'root  diminishes  their 
numerators  more  than  it  diminishes  their  denominators. 

The  negative  roots  of  all  fractions,  whether  proper  or  improper,  are, 
of  course,  algebraically  less  than  the  fractions  themselves. 

252.  In  the  foregoing  examples  the  numerators  and  denominators 
were  all  perfect  squares.  But,  if  we  have  a  fraction  whose  denomi- 
nator is  not  a  perfect  square,  we  can  readily  find  its  exact  root  to  within 
less  than  unity,  divided  by  the  denominator  of  the  fraction.  Let  it 
be  required  to  extract  the  square  root  of  |.  AYe  multiply  the  nume- 
rator and  denominator  of  the  fraction  by  the  denominator, which  does 

not  alter  the  value  of  the  fraction,  and  we  have  V I  ==  \  /  '? — -?•  = 

"        V   5  X  5 

v/4f  =  f  +.     The  root  of  the  numerator  lies  between  6  and  7.     The 

root  of  the  fraction  is  greater  than  ^,  and  less  than  |;  and  we  see  that 

I  is  the  true  value  of  the  fraction  to  within  less  than  ^.     By  this,  we 

mean  that  when  we  take  £  as  the  true  root  of  the  fraction,  we  commit 

an  error  less  than  i.    We  can,  of  course,  get  a  nearer  approximation  to 

the  true  value  of  the  fraction  by  multiplying  both  terms  of  the  fraction 

by  the  third,  fifth,  seventh,  or  some  odd  power  of  the  denominator.    This 

will  make  the  denominator  a  perfect  power,  and  its  root  can  be  exactly 


EXTRACTION    OF    ROOTS.  179 


r        1       m  -^  /8  X  5  X  5  X  5  /lOOO        31   ,       _. 

found.     Thus,  v  &  =  \  /  t e 5 f  =  V  /  ^^^T^  =  ^  +•    The 

'      "       V  5  X  5  X  5  X  5        V    G2o        25 

true  root  is  greater  than  |4,  and  less  than  -^'i.     Hence,  |i  diflfers  from 

the  true  root  by  a  quantity  less  than  ^'^. 

253.  When  the  denominator  and  numerator  are  both  imperfect 
powers,  as  in  the  example  just  given,  we  may  make  the  numerator  a 
perfect  power  by  multiplying  both  terms  of  the  fraction  by  the  first,  or 
some  odd  power,  of  the  numerator.     But,  in  this  case,  the  degree  of  ap- 

proximation  is  not  immediately  appai-ent.     Thus,  \/|  =  v/k 5  = 

I  approximatively.  The  true  root  is  greater  than  |,  and  less  than  |. 
The  degree  of  approximation  can  only  be  determined  by  reducing  these 
fractions  to  a  common  denominator.  We  have,  then,  ^8^  and  |ij,  and 
their  diifercnce  is  ^^tj.  Then,  ^|  differs  from  the  true  root  by  a 
quantity  less  than  ^%.  And  ||,  or  §,  also  differs  from  the  true  root 
by  a  quantity  less  than  ^%.  It  is  plain  that  the  degree  of  approxima- 
tion can  be  more  readily  determined  by  making  the  denominator  ra- 
tional than  by  making  the  numerator  rational.  It  is  even  preferable  to 
make  the  denominator  rational  when  the  numerator  is  already  so. 
though  the  process  of  making  the  denominator  a  perfect  square  make 
the  numerator  irrational.    Thus,  to  find  the  approximate  root  of  |,  place 

^^~  V  5X  5-  ^55-5+- 

254.  If  the  denominator  is  already  rational,  we  have  only  to  extract 
its  root  for  a  new  denominator,  and  write  over  it  the  approximate  root 
of  the  numerator  for  a  new  numerator. 

We  have,  then,  for  finding  the  approximate  root  of  any  fraction,  both 
terms  of  which  are  not  rational,  the  following 


EULE. 

Mahe  the  denominator  rational,  if  not  already  so,  hy  nudtiplyhuj 
loth  terms  of  the  fr action  hy  the  first,  or  some  odd  poicer  of  the  denomi- 
nator, according  to  the  degree  of  approximation  required,  so  that  the 
denominator  of  the  given  fraction  shall  he  the  square  power  of  the  de- 
nominator of  the  fraction  that  marks  the  degree  of  approximation. 
Then  extract  the  root  of  the  denominator  for  a  new  denominator,  and 
torite  over  it  the  approximate  root  of  the  numerator  for  a  neio  numera- 
tor.    Affect  the  nexo  fraction  with  the  douhle  sign,  to  indicate  that  there 


180  FOKMATION     OF     THE     TOWERS     AND 

arc  txco  roofs  equal,  n-ith  contrary  air/iis.  If  the  denominator  be 
already  rational,  and  a  greater  drgree  of  ajjproximation  is  required 
than  that  indicated  hy  unity  divided  hy  the  root  of  the  denominator, 
mvltijjly  hoth  terms  of  the  fraction  hy  the  third,  fifth,  seventh  power, 
d-c,  of  the  denominator,  according  to  the  degree  of  approximation 
required,  and  proceed  as  hefore. 

EXAMPLES. 

1.  Extract  the  square  root  of  J  to  witliin  less  than  h  of  its  true 
value.  -^ns.  ±  *. 

2.  Extract  the  square  root  of  I  to  within  less  than  i  of  its  true 
value.  ^«s.   ±  |-. 

.  3.  Extract  the  square  root  of  I  to  within  less  than  h  of  its  true 
value.  ^"s.   ±  h. 

4.  Extract  the  square  root  of  |  to  within  less  than  j'g  of  its  true 
value.  Ans.   ±  ||. 

5.  Extract  the  square  root  of  |  to  within  less  than  1  of  its  true 
value.  -Ans.   zfc  |,  nearer  4  than  |. 

6.  Extract  the  square  root  of  f  to  within  less  than  ^V  of  its  true 
value.  Ans.  ±  ^|. 

7.  Extract  the  square  root  of  -^^  to  within  less  than  ^'^^  of  its  true 
value.  Ans.   ±  ^o. 

8.  Extract  the  square  root  of  -^^  to  within  less  than  ^^^  of  its  true 
value.  Ans.   ±:  f  |fi. 

In  the  first  example,  the  multiplier  of  both  terms  of  the  fraction  is 
2;  in  the  second,  (2)^;  in  the  third,  unity;  in  the  fourth,  {-^f;  in 
the  eighth,  (27)^ 

255.  Since  the  denominator  of  the  fraction  may  be  raised  to  as  high 
an  even  power  as  we  please,  it  is  evident  that  the  degree  of  approxi- 
mation can  be  made  as  close  as  we  choose  to  the  true  value  of  the 
fraction. 

256.  We  may,  by  a  similar  process,  determine  approximatively  the 
roots  of  incommensurable  numbers  to  within  less  than  unity,  divided 
by  any  whole  number. 

Let  it  be  required  to  determine  the  square  root  of  2  to  within  less 


EXTRACTION    OF    ROOTS.  181 


than  1  of  its  true  value.     Then,  v/2  =  \ /"  ^ }^^   =  V-]|  =  |  +. 

The  true  value  lies  between  §  and  §,  and,  therefore,  |  differs  from  the 
true  value  by  a  quantity  less  than  i.  We  multiplied  and  divided  the 
given  number  by  the  square  of  the  denominator  of  the  fraction  that 
marked  the  degree  of  approximation  required.  This,  of  course,  did  not 
alter  the  value  of  the  given  number,  it  simply  placed  it  under  the  form 
of  a  fraction  with  a  rational  denominator.  The  nest  step  was  to  ex- 
tract the  root  of  the  numerator  to  within  the  nearest  unit,  and  to  write 
the  result  over  the  exact  root  of  the  denominator. 

To  demonstrate  a  general  rule  applicable  to  any  number,  and  true  for 

any  degree  of  approximation,  let  a  be  the  number,  and  —  the  fraction 

that  marks  the  degree  of  approximation.     Then,  -^"(7= -v  / ^ — . 

Let  J*  denote  the   root  of  ojtMo  within  less  than  unity;    in  other 
words,  let  r  denote  the  entire  part  of  the  root  of  an''.     We  will  then 

have  Va  =  \/-^  >  \/^,  or  -,  and  <  \/^-^,  or  -— . 

)•  ?•-+- 1 

The  true  root  of  a,  then,  lies  between  the  numbers  —  and ,  which 

n  n 

differ  from  each  other  by  — .     Hence,  —  differs  from  the  tme  root  by 

1                        ?•  +  1 
a  quantity  less  than  — .     So,  also,  differs  from  the  true  root  by 

1  r 

a  quantity  less  than  — .     We  then  have  a  riuht  to  take  either   — ,  or 

n  ^  n 
r  4-  1 
,  for  the  approximate  root.     That  one  is  taken  to  which  the  root 

lies  nearest.  , 

To  find  the  approximate  root  of  any  number,  a,  to  within  less  than 

—  of  its  true  value,  we  have  the  following 


Multiply  and  divide  the  given  numler  hy  the  square  of  the  denomi- 
nator of  the  fraction  that  marks  the  degree  of  ajyproximation.  Ex- 
tract the  root  of  the  numerator  of  the  fraction  thus  formed  to  within  the 
nearest  unit,  and  set  the  resvlt  over  the  exact  root  of  the  denominator. 
Give  the  doidjle  sign  to  the  root. 
16 


182  FORMATION     OF    THE    POWERS    AND 


EXAMPLES. 

1.  Extract  the  square  root  of  2  to  within  less  than  ^  of  its  true 
value.  ^«s-   ±  I- 

2.  Extract  the  square  root  of  50  to  within  less  than  J  of  its  true 
value.  ^i«s.  ±  2_8. 

3.  Extract  the  square  root  of  50  to  within  less  than  -J^  of  its  true 
value.  -^'Ihs.   ±  %^. 

4.  Extract  the  square  root  of  50  to  within  less  than  ^^^  of  its  true 
value.  ^ns.  ±  fM. 

It  is  obvious  that,  by  increasing  the  denominator  of  the  fraction  that 
marks  the  degree  of  approximation,  we  may  make  the  approximate  in- 
definitely near  to  the  true  value  of  the  root  of  the  given  number. 

257.  Apiiroxlviate  roots  of  wliole  numhers  expressed  dedmally. 

To  extract  the  root  of  any  whole  number,  a,  to  within  any  decimal 
<if  its  true  value,  we  have  only  to  change  the  decimal  into  an  equiva- 
lent vulgar  fraction,  J^,  ji^,  or  whatever  it  may  be,  and  then  multiply 
both  terms  of  the  fraction  by  10^  100',  &c.  Whatever  the  decimal, 
which  marks  the  degree  of  approximation,  may  be,  it  can  be  changed 

into  a  vulgar  fraction,  ^^  ;  in  which  m  is  a  positive  whole  number, 

^n-eater  by  unity  than  the  number  of  cyphers  between  the  decimal  point 
and  first  significant  figure.  Thus,  -1  =  -y— -„  and  m  =  1 ;  -01  =  -j^^' 
and  m  =  2 ;   -001  =  — y-^,  and  m  =  o. 

j_        la  {\n- 
Hence,  -J  a  =^  W  — -.   ^  ^,^  • 

Representing  the  entire  part  of  the  root  of  a  (lO)^"  by  r,  there  re- 
/«  (107">  ^        r  >•  +  ! 

Hence,   -z is  the  true  root  to  within  less  than  ;   that  is. 


(10)"'  (10)" 

— —  differs  from  the  true  root  by  a  quantity  less  than  .       Now, 

multiplying  the  given  quantity,  a,  by  (lO)^-",  is  the  same  as  annexing 


EXTRACTION     OF     ROOTS.  183 

2m  cyphers;  for,  multiplying  it  by  (10)^,  annexes  2  cyphers;  by  (lO)'', 
annexes  4  cyphers;  by  (10/,  S  cyphers,  &c.  And  dividing  the  ap- 
proximate root  found,  r,  by  (10)°"  is  plainly  the  same  as  cutting  off 
from  the  right  of  the  root  found  m  places  for  decimals,  for  to  divide 
the  root  by  (10)',  (10)^,  (10)^  is  the  same  as  cutting  off  from  the  right 
one,  two  or  three  places  of  decimals.  Hence,  to  approximate  to  the 
true  root  of  any  given  number  to  within  a  certain  number  of  decimals, 
we  have  this 

RULE. 

Annex  tioicc  as  many  cyphers  to  the  given  ninnher  as  there  are  deci- 
mal places  required  in  the  rout,  extract  the  root  of  the  number  thus  in- 
creased to  toith  in  the  nearest  unit,  and  cut  off  from  the  right  the  re- 
quired number  of  decimal  places. 


EXAMPLE.S. 

1.  Required  v/2  to  within    1.  Ans.  14. 

2.  Required  v/2  to  within   01.  Ans.  141. 

3.  Required  v/U  to  within  -001.  Ans.  1414. 

4.  Required  v/50  to  within  -01.  Ans.  7-07. 

5.  Required  V'SO  to  within  -001.  Ans.  7-071. 

6.  Required  v/"o0  to  within   0001.  Ans.  70710. 

7.  Required  ^9000  to  within  -1.  .l/is.  94-8. 

8.  Required  ^9000  to  within  -01.  Ans.  .94-86. 

9.  Required  v/ 900(7 to  within  -OQl.  Ans.  94-869. 

10.  Required  V 145  to  within  -01.  Ans.  12-04. 

11.  Required  v/T45'tp  within  -001.  Ans.  12-041. 

12.  Required  s/ 1000  to  within  -001.  .4ns.  31-622. 
1%  Required  v/lOOO  to  within  -0001  Ans.  31-6227. 

14.  Required  V  lOOOOO  to  within  -01.  Ans.  316-22. 

15.  Required  yiOOOOO  to  within  .001.  Ans.  316-227. 

Examples  12  and  14  show  that,  to  pass  from  the  root  of  any  number 
to  the  root  of  a  number  100-  times  as  great,  we  have  only  to  remove  the 
decimal  point  one  place  further  to  the  right.  The  converse  is  evi- 
dently true  also. 


184  FORMATION    OF    THE    POWEl 


MIXED  NUMBERS. 

258.  The  approximate  root  of  mixed  numbers  can  now  readily  be 
found.  Suppose  it  be  required  to  find  the  approximate  root  of  2-5  to 
witliin  -1.  If  we  annexed  two  cyphers,  as  before,  the  result  would  be 
2500  J  and  then,  when  we  shall  have  come  to  point  off"  into  periods,  the 
whole  number,  2,  will  be  united  with  the  decimal  5.  The  root  found  will 
be  5,  which  is  plainly  absurd.  But,  2-5,  changed  into  an  equivalent  vul- 
gar fraction,  is  fg.  Hence,  by  the  rule  for  vulgar  fractions,  v/2-5  = 
N^fl  =  \/y§§  =  t|  +  =  1"5  +  •  We  see,  that  in  the  present  instance, 
we  hav^  annexed  a  single  cypher,  which  made  the  decimal  places  even, 
and  double  the  number  of  places  required  in  the  root.  We  next 
pointed  off  from  the  root  the  number  of  decimal  places  required.  If  we 
are  required  to  find  the  approximate  root  of  2-5  to  within  ^i^,  yj^^,  or 

•01.  Then,  V2^  =  v/fl  -  \/'^\ll^^  =  iB§  -  1-58-  We 
have,  obviously,  added  three  cyphers  to  5,  and,  therefore,  made  the 
number  of  decimal  places  even  and  equal  to  the  number  of  places  re- 
quired in  the  root.  In  pointing  off  for  decimals,  we  have  only  pointed 
off  two  places,  the  number  required  in  the  root.  To  demonstrate 
the  rule  in  a  general  manner,  let  a  be  the  entire  part  of  the  mixed 

number,  and  ^z—;^  the  decimal  part.     Then  the  given  number  will  be 


h 

ah 

~  10"'  ■ 

Hence,  \/«' 

-\/ 

ab 
(10)"' 

V^ 

b  X  (10)™ 
(10)- 

\/' 

X  h  (10)" 
(10)'"' 

■>(ior'»"^ 

< 

r+  1 

(10)"'  • 

In   wli 

lich    r 

represents 

the  entire  part  of  the  root  of  a-X  h  (10)"'.  Now,  the  multiplication 
of  b  by  (10)""  is  the  same  as  annexing  m  eypliers  to  b,  and  whenever 
m  is  odd,  the  number  of  decimal  places  will  be  even,  and  double  the 
number  required  in  the  root.  When  m  is  even,  the  decimals  will  not 
be  mixed  with  the  whole  numbers,  as  in  the  mixed  number  3-4^  and 
there  need  be  no  cyphers  annexed. 

We  have  supposed,  in  the  general  demonstration,  that  the  number 
of  decimal  places  required  in  the  root  was  precisely  equal  to  the  num- 
ber of  decimal  places  in  the  mixed  number.  But,  if  this  were  not  the 
case,  the  denominator  of  the  equivalent  vulgar  fraction  has  only  to  be 
multiplied  by  such  a  power  of  10  as  will  make  it  10  with  an  exponent 
twice  as  great  as  the  number  of  places  required  in  the  root.     The  nu- 


EXTRACTION    OF    ROOTS.  185 

merator,  when  multiplied  by  this  power  of  10,  will  have  its  number  of 
decimal  places  even,  and  equal  to  double  the  number  of  places  required 
in  the  root. 


Annex  cyphers  until  the  mauler  of  decimal  places  in  the  mixed 
number  is  even,  and  equal  to  double  the  mimber  of  places  required  in 
the  root.  Extract  the  root  of  the  result  to  xcithin  the  nearest  unit,  and 
then  point  off  from  the  right,  for  decimals,  the  number  of  decimal 
places  required  in  the  root. 


EXAMPLES. 

1.  Extract  the  square  root  of  4-9  to  within  -1  Ans.  ±  2-2. 

2.  Extract  the  square  root  of  4-9  to  within  -01.      Ans.  rb  2-21. 

3.  Extract  the  square  root  of  4-9  to  within  -001. 

Ans.  =fc  2-213. 

4.  Extract  the  square  root  of  4-25  to  within  -01. 

Ans.  =t  2-OG. 

5.  Extract  the  square  root  of  425  to  within  -001. 

Ans.  ±2001. 

6.  Extract  the  square  root  of  9G-1  to  within  -1.        Ans.  ±  9-8. 

7.  Extract  the  square  root  of  9-Cl  to  within  -1. 

Ans.  ±3-1  exactly. 

8.  Extract  the  square  root  of  145-755  to  within  -01. 

Ans.  ±  12-07. 

9.  Extract  the  square  root  of  14575-5  to  within  -1. 

Ans.  it  120-7. 

10.  Extract  the  square  root  of  101-7  to  within  -01. 

Ans.  ±  10-08. 

11.  Extract  the  square  root  of  1001-01  to  within  -1. 

Ans.  ±  31-6. 

12.  Extract  the  square  root  of  10-0101  to  within  -01. 

Ans.  ±  3-16. 

13.  Extract  the  square  root  of  1728-555  to  within  -01. 

Ans.  ±41-57. 

14.  Extract  the  square  root  of  172855-5  to  within  -1. 

Ans.   rb  415-7. 
16* 


186  FORMATION     OF     THE     POWERS     AND 

15.  Extract  tlio  square  root  of  17285550-666  to  within  -01. 

Ans.  ±4157-58. 
IG.  Extract  the  square  root  of  1728555066-6  to  within  -1. 

Ans.  zfc  41575-8. 


ROOTS  OF  NUMBERS  ENTIRELY  DECIMAL. 

259.  Let  it  be  required  to  extract  the  square  root  of  -4.  This  deci- 
mal, changed  into  an  equivalent  vulgar  fraction,  is  j%.  Hence,  v/  -4  — 
y/-A^  =  ■v/j''g%  ^  y^o,  ^^^  <C  /o-  Hence,  -6  is  the  approximate  root 
to  within  less  than  -1. 

We  see  that  the  number  of  decimal  places  was  made  even  before  the 
root  was  extracted. 

If  we  were  required  to  extract  the  square  root  of  -4  to  within  -01, 
•001,  the  denominator  of  the  equivalent  vulgar  fraction  must  be  made 

(loy,  (io)«. 

The  decimal  fraction  can  be  written  --— . 

10™ 


/  ^'  n^  X  (10)-  ^       r  ,  ^r  +  1 

Hence,  s/ ^^  =  \/  -^j^jy^  >  ^^.,  and  <  --. 

It  is  plain  that  the  multiplication  of  h  by  (10)""  makes  the  number 
of  decimal  places  even.  In  pointing  off  for  decimals,  as  many  places 
must  be  cut  off  from  the  right  as  there  are  periods. 


RULE. 

Aiinex  cyphers  to  the  given  decimal  until  its  places  are  even.  Ex- 
tract the  root  of  the  result,  as  in  ivhole  numbers,  and  cut  off  from  the 
right,  for  decimals,  as  man ij  places  as  there  are  periods  in  the  numher 
ivhose  root  was  extracted.  If  it  he  required  to  extract  the  root  to 
within  a  certain  decimal,  annex  cyphers  until  the  numher  of  periods  is 
equal  to  the  numher  of  places  required  in  the  root. 

EXAMPLES. 

1.  Required  V"^,  Ans.  db  -5  exactly. 

2.  Required  %/ -9  to  within  -1.  Ans.  =b  -9t 

3.  Required  V"-09.  Ans.  d=  -3  exactly. 

4.  Required  v/-009  to  within  -01.  Ans.  ±  -09. 


EXTRACTION    OF    ROOTS.  187 

5.  Kequired  v/-0009.  Ans.   ±z  -03  exactly. 

6.  Required  v/-725.  Aus.  d=  -85  exactly. 

7.  Required  n/-U725  to  within  -01.  --I'^s.  ±  -27. 

8.  Required  v/ -00725  to  within  -001.  Ans.  ±  -085. 

9.  Required  V~^.  Ans.  ±  -9  exactly. 

10.  Required  v/-00«l.  Ans.  rb  -09  exactly. 

11.  Required  v/ -00081  to  within  -001.  Ans.   ±  -028. 

12.  Requtred  V"^000081.  Ans.   =b  -009  exactly. 

13.  Required  V  0000081  to  within  -0001.  Ans.  ±  -0028. 

14.  Required  v/-0U4.  Ans.  ±  -12  exactly. 

15.  Required  V  -04937284.  Ans.  ±  -2222  exactly. 

16.  Required  V  -05555555  to  within  .0001.  Ans.  ±  -2357. 

17.  Required  V -1111088889.  Ans.  ±  -33333  exactly. 

These  examples  show  that  the  roots  of  decimals  are  greater  than  the 
decimals  themselves.  This  ought  to  be  so,  for  all  purely  decimal  num- 
bers can  be  changed  into  proper  fractions. 

SQUARE  ROOT  OF  FRACTIONS  EXPRESSED  DECIMALLY. 

260.  Vulgar  fractions  may  be  changed  into  decimal  fractions,  and 
then  their  roots  may  be  extracted  by  the  last  rule.  If  the  given  frac- 
tion be  mixed,  it  must  be  first  reduced  to  an  improper  fraction  and 
then  changed  into  an  equivalent  decimal  fraction. 

EXAMPLES. 

1.  Required  V'J  =  %/ -6606  =  +  -81.  Ans.  d=  -81. 

2.  Required  y/~i  =  V'^-  Ans.  =t  -5  exactly. 

3.  Required  v^^  =  v^llllllll.  Ans.   ±    3333. 

4.  Required  ^~^  =  v/-0625.  Ans.  ±  -25  exactly. 

5.  Required  J^j  =  s/  012345079012.  Ans.   =b  -111111. 

6.  Required  VT  to  within  -01.  Ans.  ±  -86. 

7.  Required  V2|^5  to  within  -01.  Ans.   ±  158. 

8.  Required  v/2l.  Ans.  ±15  exactly. 

9.  Required  y  J~  to  within  -001.  Ans.  d=  -301. 


188  FORMATION    OF    THE    POWERS    AND 

10.  Required  Vl2f^  to  within  -001.  Ans.  ±  3-523. 

11.  Required  V251  to  within  -001.  Ans.  ±  5-011. 

12.  Required  V'S^^  to  within  -01.  Ans.  ±  2-25  exactly. 

The  foregoing  examples  show  that  a  vulgar  fraction,  which  is  a  perfect 
square,  may  or  may  not  have  an  exact  root  when  changed  into  a  deci- 
mal fraction. 

261. — GENERAL   EXAMPLES.  » 


1.  Required  V48303584-4856  to  within  -001. 

A71S.  do  0950-078. 

2.  Required  ^  25012001-44  to  within  -1. 

Ans.  ±  5001-2  exactly. 


3.  Required  ^-0289  to  within  -01.  Ans.  ±  -17  exactly. 

4.  Required  V -000144  to  within  -001.  Ans.  ±  -012  exactly. 

5.  Required  V^^  to  within  -001.  Ans.  ±  -242. 

6.  Required  ^/^  to  within  -001.  Ans.  ±  -125  exactly. 

7.  Required  v/^|^  to  within  -0001.  Ans.  ±  -0G25  exactly. 

8.  Required  v/^oW  *«  ^i*^^""  -00001. 

A71S.  ±  -03125  exactly. 


9.  Required  V1728  to  within  -001.  Ans.  ±  41-569. 

10.  Required  v/1728  to  within  jl^.  Ans.  ±  6j?^8_6. 

11.  Required  V16|l|  to  within  -001.  Ans.  ±  4-103. 

12.  Required  v/^'^  to  within  -1.  Ans.  ±  -2  exactly. 

13.  Required  s/~^  to  within  -00001.  Ajis.  db  -03703. 

14.  Required  V^j  to  within  -00001.  Ans.  ±  -01234. 


EXTRACTION  OF  THE  CUBE  ROOT  OF  NUMBERS. 

262.  The  cuhe  or  third  power  of  a  quantity  is  the  product  arising 
from  taking  the  quantity  three  times  as  a  factor.  Thus,  the  cube  of 
m  is  m  X  m  x  m  =  m".  The  cube  root  of  a  quantity  is  one  of  the 
three  equal  factors  into  which  the  quantity  can  be  resolved.  The  pro- 
cess of  extracting  the  cube  root  consists,  then,  in  seeking  one  of  the 
equal  factors  which  make  up  the  given  quantity.  When  the  quantity 
can  be  exactly  resolved  into  its  three  equal  factors,  it  is  said  to  be  a 


EXTRACTION    OF    ROOTS.  189 

perfect  cube.  But,  when  one  of  its  factors  can  only  be  found  approxi- 
uiatively,  it  is  said  to  be  an  incommensurable  quantity,  or  an  imperfect 
cube.  Thus,  ^S,  and  ^^27  are  perfect  cubes;  but  -i/y,  and  </2i5  are 
incommensurable,  or  imperfect  cubes. 

The  first  ten  numbers  are 

1,  2,    3,     4,      5,      6,      7,      8,      9,      10. 
and  their  cubes 

1,  8,  27,  64,  125,  216,  343,  512,  729,  1000. 

Reciprocally,  the  numbers  of  the  first  line  are  the  cube*  roots  of  the 
numbers  of  the  second  line. 

We  see,  by  inspection,  that  there  are  but  nine  perfect  cubes  among 
all  the  numbers  expressed  by  one,  two,  and  three  figures.  All  other 
numbers,  except  the  nine  written  above,  expressed  by  one,  two,  or 
three  figures  will  be  incommensurable,  and  their  roots  will  be  expressed 
by  whole  numbers  plus  irrational  parts,  which  can  only  be  determined 
approximatively.  Thus,  the  ^d  consists  of  the  whole  number  2,  plus 
an  irrational  number.  Because  9  lies  between  8,  whose  cube  root  is  2, 
and  27,  whose  cube  root  is  3.  By  a  course  of  reasoning  similar  to  that 
already  employed  (Art.  248),  it  can  be  shown  that  the  cube  root  of  an 
imperfect  cube,  as  9,  cannot  be  expressed  by  an  exact  vulgar  fraction. 

For,  if  it  can,  let  —  be  that  vulgar  fraction  :  then,  ^'Wz=—,   or   9   =: 
6  ^  6 

-J.     But,  if—  be  an  irreducible  fraction,  from  what  has  been  shown, 

0  0 

7T-  must  be  an  irreducible  fraction,  and  we  then  have  a  whole  number 

equal  to  an  irreducible  fraction,  which  is  absurd.  And,  since  the 
generality  of  the  reasoning  has  not  been  affected  at  all  by  the  selection 
of  the  particular  number,  9,  we  conclude  that  the  root  of  an  imperfect 
cube  cannot  be  determined  exactly. 

Before  demonstrating  a  rule  by  which  the  roots  of  perfect  cubes  can 
be  determined,  or  the  roots  of  imperfect  cubes  found  approximatively, 
it  will  be  necessary  to  examine  the  manner  in  which  a  cube,  or  third 
power,  is  formed.  When  the  number  contains  less  than  four  figures, 
its  cube  root,  or  the  entire  part  of  that  root,  must  be  found  among  the 
first  nine  figures.  When  the  number  contains  more  than  three  figures, 
its  root  must  be  made  up  of  a  certain  number  of  tens  and  units.  Let 
a  =  tens  of  the  root,  and  b  =  the  units  of  the  root.  Then,  the  number 
will  be  (a  -f  ly.     And,  by  actually  lijultiplyiug  (a  +  l>)  the  required 


190  FORMATION    OF    THE    POWERS    AND 

number  of  times,  we  will  get  (a  +  bf  =  a^  +  Sa^b  -f  Sab^  +  i^. 
l''hat  is,  a  number,  whose  root  is  made  up  of  tens  and  units,  is  equal  to 
the  cube  of  the  tens,  plus  three  times  the  square  of  the  tens  by  the  units, 
plus  three  times  the  product  of  the  tens  by  the  square  of  the  units,  p>his 
the  ctibe  of  the  units. 

The  formula  may  be  written  (a  +  bf  =  a^  +  (Sa^  +  Sab  +  V^h. 
And  we  see  that  the  first  thing  to  be  done  is  to  extract  the  cube  root 
of  a",  then  the  tens  (represented  by  a)  will  be  known.  It  is  obvious, 
too,  that  the  true  divisor  of  the  remainder,  after  a^  has  been  taken  out, 
to  find  b,  or  the  units,  is  the  coefficient  of  b,  (3a^  +  Sab  -f  V^) ',  but. 
since  b  is  unknown,  the  last  two  terms  of  this  coefficient,  Sab  and  //, 
are  unknown.  Hence,  we  are  compelled  to  use  Sa^  as  an  approximate 
divisor  of  the  first  remainder  to  find  b.  Now,  b,  thus  found,  will,  in 
every  case,  be  too  great,  because  the  divisor  has  been  too  small ;  but  it 
may  happen  that  when  b  is  small,  the  addition  of  Sab  +  b^  to  Sa^  would 
)iot  diminish  the  quotient  b  by  unity,  and  then  there  will  be  no  error 
committed  in  assuming  b  to  be  the  true  quotient,  provided  we  increase 
our  divisor  by  Sab  -\-  b^,  and  form  the  parts  that  enter  into  the  first 
remainder. 

We  will  illustrate  by  an  example.  Let  it  be  required  to  extract  the 
cube  root  of  1728. 

We  begin  by  separating  the  three  right  hand  figures,  because  the 

(I  +  b  cube      of 

Srt^         =300     l'72S  12  the      tens 

(Sa -\-b)b  =  (SO  i- 2)2=    64     1000=  a^  must     be, 

(Sa'  +  Sab+P)  =364        72S  =  (Sa'  + Sab +  b')b     ^^      ^^''^^' 

728=(Sd^  +  Sab  +  b^)b  thousands, 
and,  there- 
fore, cannot  be  contained  in  the  right  hand  period.  We  next  seek  the 
greatest  cube  in  the  left  hand  period,  which  is  really  1000 ;  the  root  is 
one  ten,  or  10,  we  set  it  on  the  right,  after  the  manner  of  a  quotient  in 
division.  Sa^,  or  300,  is  assumed  as  an  approximate  divisor  of  the 
remainder  after  a*,  or  1000,  has  been  taken  from  1728.  The  remain- 
der, 728,  being  represented  by  (3(t'^  +  Sab  +  li^)b,  the  true  divisor  to 
find  b  is,  of  course,  the  parenthetical  coefficient  of  b.  Having  found  b, 
or  2,  by  means  of  the  approximate  divisor,  we  set  it  on  the  right  of  the 
ten,  separating  it  by  a  point  to  indicate  that  it  is  of  a  different  denomi- 
nation. We  next  add  {Sa  +  b)b,  or  64,  to  Sa^,  or  300,  and  we  have 
364  for  the  true  divisor,  provided  that  b  has  been  found  correctly. 
This  divisor  we  will  call  the  supposed  true  divisor.     Finally,  we  multi- 


EXTRACTION     OF     KOOTS.  191 

ply  (3a^  +  Sah  +  ¥,  the  supjjosed  true  dioisor  by  h,  and  the  product 
made  up  (3a^  +  3a6  +  i^)i,  the  parts  entering:  into  the  remainder. 
The  product  thus  formed  being  exactly  equal  to  the  remainder,  proved 
two  things :  first,  that  if  the  true,  instead  of  the  approximate,  divisor 
had  been  used,  h  would  not  have  been  diminished  by  unity;  and, 
second,  that  the  number  1728  is  an  exact  cube,  and  that  its  root  is 
12.  Iq  the  present  example,  it  is  easy  to  see  why  b  was  found 
correctly  by  using  the  approximate  instead  of  the  true  divisor ;  a  and 
h  being  both  small,  the  omission  of  ^ah  +  V^  did  not  materially  aifect 
the  divisor. 

\V'e  have,  from  the  foregoing,  a  simple  test  by  which  to  ascertain 
whether  h,  found  by  using  the  approximate  divisor,  3o^,  must  be  di- 
minished. "Whenever  the  supposed  true  divisor  will  give  a  less  quo- 
tient than  I,  we  conclude  that  h  was  too  great,  and  it  must  be  dimi- 
nished by  1,  2,  &c.,  until  the  supposed  true  divisor  will  enter  into  the 
remainder  the  same  number  of  times  as  the  approximate.  It  will  never 
happen  when  the  unit  figure  of  the  root  is  small  with  respect  to  the 
tens,  as  in  71,  82,  93,  &c  ,  that  the  h,  found  by  using  3a^  as  a  divisor, 
must  be  diminished  by  unity,  or  some  greater  number.  But,  when  the 
unit  figure  is  great  with  respect  to  the  tens,  as  in  18,  19,  &c.,  it  may 
be  necessary  to  diminish  h  by  one  or  more  units.  The  reason  of  this 
is  plain.  We  will  illustrate  by  an  example.  Kequired  the  cube  root 
of  5832. 

"We  see  that  i,  found  by  using  30^^  as  a  divisor,  is  IG,  and  the  sup- 
posed   true    divisor    will 

then  be  1036;    but  this,  5832  ] 

instead    of    entering    IG  3a'  =r      300  1000 

times  into  the  remainder,     (30  +  16)16  =      736  4832     ^' 

4832,  will   only  enter  4  (1036)16  =  16576 

times.     Besides,  1036,  or 

the  supposed  (3a^  +  ^ah  +  b^),  when  multiplied  by  b,  will  give  a  pro- 
duct, 16576,  greater  than  the  remainder.  The  unit  figure  is  then  too 
great,  and  has  to  be  diminished  by  8 ;  this  diminution  can  only  be 
determined  by  trial.     The  process  ought  to  have  been 


5832    a  +  b 

1000     1'8 

3a2      =  300 

4832  =  (3a2  +  3o6  -f  b^)h 

(30  -f  8)  8  =  (3«  +  h)h  =  304 

4832  =  (3a2  -|-  ^ab  +  V)h 

3a2  +  3a5  +  h^                =604 

192  FORMATION     OF    THE    P  0  AV  E  R  S    AND 

The  true  divisor  by  trial  was  found  to  be  604.     This  divisor  was  repre- 
sented by  (3a^  +  Sah  +  J/),  and,  when  multiplied  by  b,  or  8,  gave 
(Sa^  +  Sab  +  V)b,  the  remainder. 
Take,. as  another  example,  G859. 

6  859       I  a-\-b 
3a^  =300     1000       I     19 

(3a  +  &)i  =  (30  +  9)9  =  351      5  859  =  (3a^  +  Sab  +  h^j 
3a2  +  Sah  -^W  =  651      5  859  =  (3a^  +  Sab  +  W)b 

By  trial,  9  was  found  to  be  the  second  figure  of  the  root,  the  sup- 
posed true  divisor  was  then  determined  to  be  651,  and  this,  multiplied 
by  b  or  9,  gave  5859.  The  given  number  was  then  a  perfect  cube,  and 
19  its  exact  root.  In  this  case,  the  supposed  true  divisor  is  actually 
the  true  divisor. 

263.  The  process  for  extracting  the  cube  root  of  a  number  below 
10000  may  be  without  any  difficulty  extended  to  all  numbers  whatever. 
Suppose  the  number  to  be  1881365963625,  its  root  may  still  be  re- 
garded as  made  up  of  tens  and  units,  the  cube  of  the  tens  cannot  enter 
into  the  last  three  figures,  625,  on  the  right,  and  they  may,  therefore, 
be  separated  from  the  other  figures.  The  greatest  cube  contained  in 
1881365963  must  have  more  than  one  figure  in  its  root,  because  the 
number  is  greater  than  1000,  which  contains  two  figures  in  its  root. 
Then  the  root  of  1881365963  may  be  regarded  as  made  up  of  tens  and 
units ;  and,  as  the  cube  of  the  tens  cannot  give  a  less  denomination 
than  thousands,  the  tens  cannot  be  found  in  the  last  three  figures,  963, 
which  may,  therefore,  be  separated  from  the  other  figures.  After  the 
separation  of  963  the  foregoing  reasoning  may  be  repeated,  and  thus 
dividing  the  number  into  periods  of  three  figures  until  we  come,  at 
length,  to  the  place  occupied  by  the  cube  of  the  tens  of  the  highest 
order.  The  period  on  the  left  thus  found,  may  contain  but  one  figure, 
as  in  the  present  example,  or  it  may  contain  two  figures,  or  even  three 
figures.  We  will  designate  the  tens  of  the  highest  order  by  a',  those 
of  the  second  order  by  a",  those  of  the  third  by  a"',  &c.  In  like  man- 
ner, we  will  designate  the  units  of  the  highest  order  by  U,  those  of  the 
second  order  by  b",  &c. 


EXTRACTION    OP    ROOTS.  193 

Sa'ariSOOir  1st  Ap.  divisor.     1881365*963*625  I  l'2'3*4*5 

(30+2)2=  &i=(Sa'+b')b'  1000 _| 

3a'a+3a'6'+6'2=:3B4=True  divisor.  SSI— (Za'^+3a'b' +1/^)1/+ 

64=(3a'+i/)//  728=(3a'^+3a'6'+6'2)6' 

4=6'^  153  365=(.3a"2-j-3a"?y'+fr"a)6"-i- 

3a"!»=43200=2d  Ap.  divisor.       132  ^Q~=(5a"^-{-3a"h"^+b''^)V 
(360  -f3)3=  10%9=(Za"+h'')b"  20  498  365=(3a"'a+3a"'V"+i/"3)6"'+ 

44289=True  divisor.  18213904=(3a"'a+3a"'i/"+^"')6"' 

10S9=(3a"+i"j^"  2  285  059  625=  (3a""»+3<i""J/"'+6"">)6"" 

9= h""^  2JS5^59625=(3a"">+3a""&""+6""»)  V" 

3ri"''=4.o38T00=3d  Ap.  divLsor. 
(3090+4)4=     14-76=<3a"'+6"0V" 
4553476=True  divisor. 
14776=(3a"'+6"')'/" 
16=6'"^ 
(3a''')»=45O>)20S0O=4th  .\p.  divisor 
(37026+5)5=      185125 

45701  ly25=True  divisor. 

lu  this  example,  we  begin  by  dividing  the  nuuiber  off  into  periods 
of  three  figures  each ;  the  period  on  the  left  contains  but  one  figure. 
We  extract  the  greatest  cube  contained  in  this  period,  and  set  the  root, 
1,  on  the  right.  We  then  have  found  a',  or  the  tens  of  the  highest 
denomination.  We  then  bring  down  the  next  period.  Now,  since  but 
two  periods  are  under  consideration,  the  a',  or  1,  found,  will  be  tens 
with  respect  to  these  two  periods,  and  therefore,  three  times  its  square, 
or  3a'^  =  300.  This  is,  then,  the  first  approximate  divisor.  Dividing 
881  by  it,  we  get  2,  or  the  supposed  units  contained  in  the  root  of  the 
two  left  hand  periods,  we  then  add  64,  or  (Sa'  +  ?/)6',  to  the  approxi- 
mate divisor,  and,  if  U  be  the  true  units  of  the  root,  364,  or  3a'^  -f 
SaU  +  h'^,  will  be  the  true  divisor.  Now,  since  364  enters  into  881, 
the  same  number  of  times  that  300  does,  b'  has  been  found  correctly. 
We,  therefore,  multiply  364  by  2,  and  we  form  the  three  parts,  (Sa'^ 
+  'dab'  -\-  b'^)b',  of  which  the  remainder,  881,  is  composed,  except 
some  tens  and  units  which  have  been  incorporated  in  it  from  the  cube 
of  the  tens  of  a  lower  denomination.  We  subtract  728  from  881,  and 
bring  down  the  next  period.  Now,  the  next  approximate  divisor  is 
plainly  three  times  the  square  of  120,  or  43200.  But  the  shortest  way 
of  getting  three  times  the  square  of  120  is  to  add  64,  and  the  S({uare 
of  b',  or  2,  to  the  true  divisor,  and  multiply  the  sum  of  the  three  by 
100.  The  reason  of  this  is  apparent.  The  tens  of  the  next  denomina- 
tion is  found  by  dividing  153365  by  43200 )  the  quotient  3  is  set  on 
the  right  of  the  2  in  the  root,  separated  from  it  by  a  dot.  We  next 
add  1089,  or  (3a"  +  b")b",  to  43200,  and  we  have  the  second  true 
divisor.  Multiplying  this  divisor  by  3,  and  subtracting  the  product 
from  153365,  and  bringing  down  the  next  period,  we  have  a  new 
17  N 


194  FORMATION    OF    THE    POWERS    AND 

number,  the  root  of  whose  unit  is  to  be  found.  The  approximate  di- 
visor to  find  h'"  must  be  three  times  the  square  of  1230 ;  and  this  is 
most  readily  found  by  adding  1089  and  the  square  of  h",  or  3,  to 
43200,  and  multiplying  the  sum  by  100 ;  that  is,  annexing  two  cyphers 
to  the  sum.  The  approximate  divisor  thus  found,  4538700,  enters  4 
times  in  20498365,  and  the  true  divisor,  when  formed,  enters  the  same 
number  of  times ;  4  is  then,  truly,  the  fourth  figure  of  the  root.  Mul- 
tiply 4538700,  the  true  divisor,  by  the  last  figure  of  the  root,  subtract 
the  product  from  20498365,  and  bring  down  the  next  period.  The 
next  approximate  divisor  is  three  times  the  square  of  12340 ;  this  can 
readily  be  formed  like  the  preceding.  The  true  divisor  is  formed  when 
5,  the  final  unit,  has  been  determined.  The  true  divisor,  multiplied 
by  5,  gives  a  product  equal  to  the  last  remainder.  Hence,  the  given 
number  is  a  perfect  cube,  and  12345  is  its  exact  root. 

The  reason  of  the  above  process  becomes  very  plain  upon  a  slight 
examination.  It  was  established  at  the  outset,  that  the  tens  and  units 
contained  in  the  first  two  periods  could  be  sought,  independently  of  the 
other  periods.  Having  found  found  1  ten  and  2  units  in  the  first  two 
periods,  it  is  plain  that  this  number,  12,  is  the  tens  in  123,  the  root 
found  in  the  first  three  periods.  So,  123  may  be  regarded  as  the  tens 
of  the  first  four  periods.  The  root  found  in  these  periods  is  1234, 
made  up  of  123  tens  and  4  units.  In  like  manner,  1234  may  be  re- 
garded as  the  tens  of  the  root  of  the  whole  five  periods.  In  fact,  the 
root  found,  12345,  is  made  up  of  1234  tens  and  5  units. 

We  then  had  tens  of  different  denominations,  1  ten,  12  tens,  123 
tens,  and  1234  tens.  To  apply  the  general  formula,  (a  -f  lif  =  a'  + 
3«^&  -f  ^aW  -f  1/,  for  the  cube  of  a  number  made  up  of  tens  and  units, 
to  determining  the  tens  and  units  contained  in  any  two  consecutive 
periods,  it  became  necessary  to  distinguish  the  tens  by  dashes  to  indi- 
cate the  denomination  to  which  they  belonged. 

We  will  add  another  example,  to  show  more  fully  the  application  of 
the  preceding  principles.     Required  the  cube  root  of  997002999. 
Sa"*  =  24300  =  App.  divisor.        997'002'999  I  99  9 

(3a'  -i-  h')V  =    2511  729  I 


la"  +  h")h" 


26811  = 

True  divisor. 

268  002 

2511 

241  299 

81 

26  703  999 

2940300: 

=  2d  App.  divisor 

.  26  703  999 

26811 

2967111 

=  2d  True  divisor. 

EXTRACTION     OF    ROOTS.  195 

In  this  example,  the  cjuotient  of  the  division  of  2G8002,  by  the  ap- 
proximate divisor,  24300,  is  ]  0 ;  but  this  is  manifestly  absurd.  By 
trial,  9  is  found  to  be  the  second  figure  of  the  root,  because  the  true 
divisor,  26811,  enters  9  times,  neither  more  nor  less,  in  268002. 

We  have  for  the  extraction  of  the  cube  root  of  any  number  above 
1000,  the  foUovring 

RULE. 

I.  Begin  on  the  right  and  divide  off  periods  of  three  jigures  each. 
There  will  remain  on  the  left  a  period  of  one,  tico,  or  three  figures. 

II.  Extract  the  greatest  cube  contained  in  the  left  hand  period,  and 
set  the  root  on  the  right,  after  the  manner  of  a  quotient  in  division. 
Subtract  the  cube  of  the  root  from  the  left  hand  period,  and  brivg 
doivn  the  next  period. 

III.  Talce  three  times  the  sqnai-e  of  the  root  found,  regarded  as  tens, 
and  set  it  on  the  left  as  an  approximate  divisor;  see  how  often  this 
enters  into  the  first  remainder.  The  quotient  icill  be  the  second  figure 
of  the  root,  or  something  greater.  Add  to  the  ajyproximate  divisor  the 
product  of  three  times  the  first  figure  of  the  root,  regarded  as  tens,  plu:^ 
the  second  figure  of  the  root  by  the  second  figure  of  the  root.  The  sum 
of  this  product  and  the  approximate  divisor  %oiU  be  the  true  divisor 
if  it  enter  into  the  remainder  the  same  number  of  times  as  the  ajiproxi- 
mate  divisor.  Multiply  the  true  divisor  by  the  second  figure  of  the 
root,  sidttract  the  product  from  the  first  remainder,  and  bring  doicn  the 
next  period. 

IV.  Add  to  the  first  trtie  divisor  the  same  number  that  was  before 
added  to  the  approximate  divisor,  plus  the  square  of  the  second  figure 
of  the  root,  and  annex  two  cyphers  to  the  sum,  the  residt  will  be  the 
second  approximate  divisor.  The  true  divisor  is  found  as  before. 
Continue  this  process  until  all  the  p)eriods  are  brought  doicn  ;  then,  if 
the  last  true  divisor,  midtiplicd  by  the  fined  units  of  the  root,  gives  a 
product  exactly  equal  to  the  last  remainder,  the  given  number  is  a  per- 
fect ctdie,  and  its  exact  root  has  been  found. 


1.  Required  the  cube  root  of  9261.  Ayis.  21. 

2.  Required  the  cube  root  of  85184.  Ans.  44. 

3.  Required  the  cube  root  of  8024024008.  Ans.  2002. 

4.  Required  the  cube  root  of  1371330631.  Ans.  1111. 


196  FORMATION    OF    THE    POWERS    AND 

5.  Required  the  cube  root  of  10306070(50301.  A71S.  10101. 

6.  Required  the  cube  root  of  1307631.  Ans.  111. 

7.  Required  the  cube  root  of  1879080904.  Ans.  1284. 

8.  Required  the  cube  root  of  95306219005752.  Ans.  45678. 

9.  Required  the  cube  root  of  95306219005752000. 

Ans.  456780. 
10.  Required  the  cube  root  of  468373331006.  Ans.  7766. 

APPROXIMATE  ROOTS  OF  INCOMMENSURABLE   NUMBERS. 

264.  Let  a  represent  any  incommensurable  number — we  have  seen  that 
its  root  cannot  be  expressed  by  an  exact  fraction ;  it  may,  however,  be 

truly  determined  to  within  less  than  any  fraction,  — ,  whose  numerator 

mv"        ,      ^    r^        ,    /  (r  +  ly 
IS  unity.     I-tor  a  =  -y,  and  «  ]>  — ,  and  <^  3       ;  r  denoting  the 

entire  part  of  the  root  of  an^.     Then,  since  a  is  comprised  between 

y^3  (r  4-iy   ,  ,  r  r  4-1 

-^,  and  ^^ ^— ^,  its  root  will  be  greater  than  — ,  and  less  than . 

n  n  "■  M  n 

r     .  .  1 

Hence,  —  differs  from  the  true  root  by  a  quantity  less  than  — .     Now, 

as  —  may  be  made  indefinitely  small,  the  approximate  root  may  be 

found  as  near  to  the  true  root  as  we  please ;  the  difficulty  of  extracting 
the  root  increasing,  however,  with  the  increase  of  n. 
Hence,  we  deduce  the  following 

RULE. 

Midiiphj  and  divide  the  nnmher  hy  the  cube  0/ the  denominator  of 
the  fraction  that  marks  the  degree  of  aj^proximation  ;  extract  the  root 
of  the  new  numerator  to  within  the  nearest  unit,  and  divide  the  result 
hij  the  root  of  the  new  denominator,  lohich  will  he  the  denominator  of 
the  fraction  that  determines  the  degree  of  approximation. 


1.  Required  the  cube  root  of  4  to  within  i.  Ans.  |. 

Because  ^T==  -y^^'  =  \/~  <  f,  and  >  |.       The  true 


V  or 

V' 

Ans. 

5_3 

4   ' 

Ans. 

V 

Ans. 

V 

EXTRACTION    OF    ROOTS.  197 

root  lies,  then,  between  |  and  |,  but  is  nearer  to  |  than  to  |.     Hence, 
I  is  the  true  root  to  within  less  than  i. 

2.  Required  the  cube  root  of  80  to  within  j^^.  Ans.  f  i. 
Beea„se4.W=\y^=^y™>.i<.i.      " 

3.  Required  the  cube  root  of  90  to  within  |.  A7is.   \^. 

4.  Required  the  cube  root  of  712  to  within  J,      Ans.   ^^  nearly. 

5.  Required  the  cube  root  of  1820  to  within  I.  Ans.   \i. 
G.  Required  the  cube  root  of  200  to  within  A.       Am 

7.  Required  the  cube  root  of  2397  to  within  ]-. 

8.  Required  the  cube  root  of  1531  to  within  i. 

9.  Required  the  cube  root  of  3575  to  within  I. 

10.  Required  the  cube  root  of  3375  to  within  i. 

Ans.   Yf  or  15,  commensurable. 

11.  Required  the  cube  root  of  10G2208  to  within  J-.       Ans.  -S"^. 

CUBE  ROOT  OF  FRACTIONS. 

265.  Since  the  cube,  or  third  power,  of  a  fraction  is  formed  by 
raising  the  numerator  and  denominator  separately  to  the  third  power, 
the  cube  root  of  a  fraction  can  plainly  be  found  bj-  extracting  the  root 
of  the  numerator  and  denominator  separately.  The  root  of  the  nume- 
rator written  over  the  root  of  the  denominator  will  then  be  the  root  of 
the  fraction. 

There  are  three  cases:  1.  The  numerator  and  denominator  of  the 
given  fraction  may  be  both  perfect  cubes,  and  then  the  root  of  the  one 
written  over  the  root  of  the  other  will  be  the  required  root.  2.  The 
numerator  may  be  an  imperfect  cube,  and  the  denominator  a  perfect 
cube,  then  the  root  of  the  numerator  extracted  to  within  the  nearest  unit, 
written  over  the  exact  root  of  the  denominator,  will  be  the  approximate 
root  to  within  less  than  unity,  divided  by  the  root  of  the  denominator. 
3.  Both  the  num.erator  and  denominator  may  be  imperfect  cubes,  or 
the  denominator  only  may  be  an  imperfect  cube,  then  the  denominator 
must  be  made  a  perfect  cube  by  multiplying  it  by  its  second  power. 
The  numerator  must  also  be  multiplied  by  the  same  number,  otherwise 
17* 


19S  FORMATION     OF    THE    POWERS    AND 

the  value  of  the  fraction  will  bo  altered.  The  root  of  the  new  nume- 
rator, to  within  the  nearest  unit  written  over  the  exact  root  of  the  new 
denominator,  will  be  the  approximate  root  required.  If  a  nearer  degree 
of  approximation  be  required,  both  terms  of  the  fraction  may  be  multi- 
plied by  the  5th,  8th,  &c.  power  of  the  denominator.  The  reason  for 
making  the  denominator  rational  rather  than  the  numerator,  is  the 
same  as  that  given  in  the  explanation  of  the  principles  involved  in  ex- 
tracting the  square  root  of  fractions. 

The  rule  for  the  extraction  of  the  cube  root  of  fractions  belonging  to 
either  of  the  three  foregoing  classes,  is  as  follows  : 


RULE. 

Make  the  denominator  rational,  if  not  already  so,  hi/  multiplying 
hoth  terms  of  the  fraction  hy  the  square  power  of  the  denominator. 
Extract  the  root  of  the  new  numerator  to  within  the  nearest  unit,  and 
lorite  the  root  found  over  the  root  of  the  new  denominator,  which  will 
he  the  same  as  the  denominator  of  the  given  fraction.  If  a  nearer  de- 
gree of  approximation  he  required,  hoth  terms  of  the  fraction  may  he 
multiplied  hy  the  bth,  Sth,  &c.  powers  of  the  denom{?iator.  The  root 
(f  the  neic  fraction  loill  he  the  root  required. 


EXAMPLES. 

1.  Required  the  cube  root  of  tf^.  Ans.  |. 
For  (I)'  =  I  X  f  X  f  =  3^. 

2.  Required  the  cube  root  of  ^f  to  within  \.  Ans.  |. 

Because  (-|)^  =  |4,  and  (f)''  =  ^-^if .  Hence,  the  true  root  lies  be- 
tween I  and  I ;  and  |,  therefore,  difi'ers  from  the  true  root  by  a  quan- 
tity less  than  -i- 

3.  Required  the  cube  root  of  |  to  within  i.  Ans.  |. 

—         3  /2~~5^  

Because  y  i  =  \  /  - — —  =   -y^Q  ^  4,  and  <  4.      The  true  root 

V     5  .  5^  o3   ^    o'  --   o 

is  nearer  4  than  |,  and,  tliercrnre,  |  is  taken. 

4.  Required  the  cube  root  of  |  to  within  I.  Ans.  |. 
Because  in  ==  \/^  =  VW  >  h  a»<i  <  f • 


EXTRACTION    OP    ROOTS.  199 

5.  Required  the  cube  root  of  |  to  within  y^^.  Ans.  i4|. 

6.  Required  the  cube  root  of  ^-^  to  within  i.  Ans.   ^^. 

266.  The  cube  root  of  a  fraction  may  also  be  determined  to  within 
unity,  divided  by  any  power  of  the  denominator ;  we  have  only  to  mul- 
tiply both  terms  of  the  given  fraction  by  such  a  number  as  will  make  the 
denominator  a  perfect  third  power  of  the  denominator  of  the  fraction 
that  marks  the  degree  of  approximation,  and  then  extract  the  root  of 
the  new  fraction.  Let  it  be  required  to  extract  the  cube  root  of  '^^  to 
within  \,  then  the  given  fraction  must  have  its  numerator  and  denomi- 
nator multiplied  by  32,  in  order  to  make  the  denominator  64,  the  cube 
of  4,  the  denominator  of  the  fraction  that  marks  the  degree  of  ap- 
proximation. 

Then,  V^  =  *yi^-|^  =  irnj  <-  j^  aud  y  «. 

8.  Required  the  cube  root  of  ^.f  to  within  ^.  Ans.   ^. 


For  rV  =  \7%^  =  ^IW  >  e-  ^"^  < 


APPROXIMATE  ROOT  TO  WITIIIX  A  CERTAIN  DECIMAL. 

267.  There  are  two  cases :  the  given  number  whose  root  is  to  be 
found  may  be  entire,  or  it  may  be  mixed ;  partly  entire  and  partly 
decimal. 

CASE  I. 

Approximate  root  oftoJiole  numbers  to  ivitJiin  a  certain  decimal. 

If  the  decimal  fraction  that  marks  the  degree  of  approximation  be 
changed  into  an  equivalent  vulgar  fraction,  the  cube  of  its  denominator 
will  be  unity,  followed  by  three,  six,  nine,  twelve,  or  some  multiple  of 
three  cyphers.  In  other  words,  the  cube  of  the  denominator  will  con- 
tain unity,  followed  by  as  many  periods  of  three  cyphers  each,  counting 
from  the  right,  as  there  are  decimal  places  in  the  fraction  of  approxi- 
mation.  Thus  (-1/  =  {-.^J  =  ^J,^  .  (-Ol)'  =  (^1^^  =  y.^^^^. 
Then,  to  multiply  the  given  number  by  the  cube  of  the  denominator  of 
the  decimal  changed  into  a  vulgar  fraction  is  nothing  more  than  an- 
nexing three,  six,  nine,  or  some  multiple  of  three  cyphers.  After  a 
now  number  has  thus  been  formed,  the  extraction  /of  the  root  is,  of 


200  FORMATION    OF    THE    POWERS    AND 

course,  performed  just  as  when  the  fraction  of  approximation  was  a 
vulgar  fraction.  Let  it  bo  required  to  extract  the  cube  root  of  5  to 
within  -01. 


^  3/5.(100)3         V^OOOOOO.      ,^^       J/,,, 

Then,  V.  =  N/-^/  =  \/imr  >  ^^«'  '""^  <  T^^- 


RULE. 

Annex  to  the  given  numher  three  times  as  many  cyphers  as  there  are 
decimal  places  required  in  the  roof.  Extract  the  root  of  the  neio  num- 
her thus  formed  to  within  the  nearest  unit,  and  point  off  from  the  right 
the  required  numher  of  decimal  places ;  tchich  amounts  to  the  same 
thing  as  dividing  the  root  of  the  new  numher  hy  the  denominator  of  the 
fraction  of  approximation  changed  into  an  eQuivalent  vulgar  fraction. 


EXAMPLES. 

1.  Eequired  the  cube  root  of  60  to  within  -1  Ans.  3-9. 

2.  Required  the  cube  root  of  1775  to  within  -1.  Ans.  12-1. 

3.  Required  the  cube  root  of  9  to  within  -01.  Ans.  2-08. 

4.  Required  the  cube  root  of  9  to  within  -0001. 

Ans.  2-0801  nearly. 

5.  Required  the  cube  root  of  18G4967  to  within  -1. 

Ans.  123-1. 

CASE  II. 

268.  Approximate  roots  of  mixed   numhers    to    within    a   certain 
decimal. 

Let  it  be  required  to  extract  the  cube  root  of  2o  to  within  -1. 


Then,  y  2^ 


3/23  .  10^  3/2300  ^    ,  „        ^   ^  , 


Wc  see  that  when  the  decimal  has  been  changed  into  a  vulgar  frac- 
tion, and  the  denominator  made  rational,  the  numerator,  2300,  is  the 
given  number,  with  the  point  omitted,  and  with  cyphers  enough  an- 
nexed to  make  the  number  of  places,  counting  from  where  the  point 
was,  a  multiple  of  three.  If  it  had  been  required  to  extract  the  cube  root 
of  2-3  to  within  -01,.  it  would  have  been  necessary  to  add  five  cyphers 


EXTRACTION     OF     ROOTS.  201 

to  3 ;  the  number  of  decimal  places  then  would  have  been  made  a  mul- 
tiple of  8.  Now,  in  pointing  off  for  decimals,  it  is  plain  that  we  point 
off  as  many  places  for  decimals  as  there  are  places  required  in  the  root. 
For,  if  one  decimal  place  be  required  in  the  root,  the  denominator  of 
the  root  of  the  equivalent  Aiilgar  fraction  will  be  10;  if  two  places  be 
required,  it  will  be  100,  &c. 

RULE. 

Annex  cyphers  to  the  decimal  part  of  the  mixed  number  until  there 
are  three  places,  if  one  place  he  required  in  the  root ;  six  places  if  two 
he  required  in  the  root,  &c.  Extract  the  root  of  the  new  number  thics 
formed  as  a  whole  number,  and  point  off  from  the  right  the  required 
number  of  decimals. 

EXAMPLES. 

1.   Re(|uired  the  cube  root  of  2-12  to  within  -01.  Ans.  1-28. 


3/212  3/212x100^  3/2120000^,, 

For,  VTT2  =  ^—  =  V -Too^  =  N/ "100^  >  •  "< 
or  1-28. 

2.  Required  the  cube  root  of  4-1  to  within  -1.  Ans.  1-6. 


3/41  X  10^         3/4100  ^    , ,        ,  /  , , 
For,  yCT=  V^y  =  ^___  =  y^__  >  m,  and  <  jg. 

3.  Required  the  cube  root  of  888  to  within  -01.  Ans.  2-07. 

4.  Required  the  cube  root  of  68-64  to  within  -1.  Ans.  4-1. 

5.  Required  the  cube  root  of  1770-25  to  within   1.        Ans.  12-1. 

6.  Required  the  cube  root  of  1150-455  to  within  -1. 

Ans.  10-4  nearly. 

7.  Required  the  cube  root  of  5011-125  to  within  -1.      Ans.  17-1. 

In  the  4th  and  5th  examples,  one  cypher  only  had  to  be  annexed. 
In  the  last  two  examples,  no  annexation  was  required.  But  if  the 
root  in  the  last  two  examples  is  to  be  determined  within  -01,  then 
three  additional  cyphers  must  be  annexed ;  if  within  -001,  six  addi- 
tional cyphers,  &c. 

The  reason  for  annexing  cyphers  until  the  decimal  places  can  be 
separated  into  periods  of  three  figures,  is  evident,  even  without  ehang- 


202  FORMATION    OF    THE    POWERS    AND 

ing  the  decimal  into  an  equivalent  vulgar  fraction.  For  each  period 
of  three  figures  must  give  one  figure  in  the  root ,  if,  then,  the  decimal 
places  were  not  made  multiples  of  three,  when  we  come  to  point  ofi" 
from  the  right  the  decimals  would  be  mixed  with  the  whole  number. 
Suppose  we  were  required  to  find  the  root  of  8-72  to  within  -1.  Now, 
8-72  is  but  little  greater  than  8,  whose  root  is  2.  Hence,  the  root  of 
8-72  ought  to  be  but  little  greater  than  two;  but  if  we  annexed  no 
cypher  to  72,  we  would  have  but  one  period,  872,  and  the  root  would 
be  9  approximatively. 


APPROXIMATE  ROOT  OF  DECIMAL  FRACTIONS  TO  WITHIN 
A  CERTAIN  DECIMAL. 

269.  A  decimal  fraction  changed  into  an  equivalent  vulgar  fraction 
will  have  a  rational  denominator  when  its  number  of  decimal  places  are 
multiples  of  three. 

Thus,  -021  =  ^§1^,  -007681  =  toVoUo.  -123456789  =  j^^WVo- 

Hence,  if  cyphers  be  annexed  to  the  decimal  until  the  number  of 
places  are  made  multiples  of  three,  the  new  fraction,  when  changed 
into  an  equivalent  vulgar  fraction,  will  have  a  rational  denominator. 
And,  since  every  period  of  three  figures  gives  one  figure  in  the  root, 
cyphers  must  be  annexed  until  the  number  of  periods  of  three  figures 
is  exactly  equal  to  the  number  of  places  required  in  the  root. 

RULE. 

Annex  cyphers  to  the  given  decimal  until  it  can  he  divided  off  into 
as  many  periods  of  three  jigures  each  as  there  are  places  required  in 
the  root.  Extract  the  root  of  the  oicw  decimal  thus  formed  to  vnthin 
the  nearest  unit,  and  point  off  from  the  right  the  required  number  of 
decimal  places. 

EXAMPLES. 

1.  Required  the  cube  root  of  -8  to  within  -1.  Ans.  -9. 

3/800 

For,  ^.S=^-,%  =  s^  j^3-  >  /„  and  <  fo. 

2.  Required  the  cube  root  of  -08  to  within  -1.  Ans.   -4. 


For,  ^■i)>i  =  ^j^^=V^n^>j'„  and<Y\5. 


EXTRACTION     OF    ROOTS.  203 

3.  Required  the  cube  root  of  -008  to  within  -1. 

Ans.  -2,  commensurable. 
For,  </"^0U8  =  ^7^=  rt  =  -^ 

4.  Required  the  cube  root  of  -8  to  within  -01.  Ans.   "92. 

5.  Required  the  cube  root  of  -8  to  within  -001.  Ans.  -928. 
G.  Required  the  cube  root  of  -68  to  within  -1.  Atis.  -8. 

7.  Required  the  cube  root  of  -68  to  within  -01.  Ans.   -87. 

8.  Required  the  cube  root  of  -068  to  within  -1.  Ans.   4. 

9.  Required  the  cube  root  of  -0999  to  within  -01.  Ans.  -46. 

10.  Required  the  cube  root  of  -999  to  within  -01.  Ans.  -99. 

11.  Required  the  cube  root  of  0125  to  within  -01.  Ans.   -23. 

12.  Required  the  cube  root  of  -125.         Ans.   -5,  commensurable. 

In  some  of  the  examples,  it  was  necessary  to  annex  one  cypher,  in 
others,  two  cyphers;  in  others,  again,  three  cyphers.  It  will  be  seen 
that  the  root  is  greater  than  the  number ;  this  ought  to  be  so,  since  the 
denominator  of  the  decimal  changed  into  a  vulgar  fraction  is  greater 
than  the  numerator. 


APPROXIMATE  ROOT  OF  VULGAR  FRACTIOXS  TO  WITIIIX  A 
CERTAIN  DECIMAL. 

270.  Any  vulgar  fraction  may  be  changed  into  an  equivalent  deci- 
mal fraction  by  annexing  cyphers  to  the  numerator,  and  dividing  the 
new  numerator  thus  formed  by  the  denominator.  The  equivalent  deci- 
mal will  be  mixed,  or  purely  decimal,  according  as  the  given  fraction 
is  improper  or  proper.  The  division  of  the  new  numerator  by  the 
given  denominator  must  be  continued  until  the  decimal  part  of  the 
quotient  contains  as  many  periods  of  three  figures  as  there  are  places 
required  in  the  root. 


Change  the  vulgar  fraction  into  an  equivalent  decimal  fraction,  and 
make  the  decimal  part  of  the  quotient  contain  as  many  periods  of  three- 
figures  as  there  are  2:)laces  required  in  the  root.  Then  extract  the  root 
according  to  the  j^receding  rides. 


204  FOllMATION     OF     THE     POWERS     AND 

EXAMPLES. 

1.  Ecqiiired  the  cube  root  of  |  to  witlain  -1.  Ans.   -6. 

2.  Kequired  the  cube  root  of  |  to  within  -1.  Ans.  1-1. 

3.  Required  the  cube  root  of  |  to  within  -01.      Ans.  1-17  nearly. 

4.  Required  the  cube  root  of  ^  to  within  -01.  Ans.  -82. 

5.  Required  the  cube  root  of  ^^  to  within  -001.  Ans.   -436. 

General  Rcmca-Jcs  on  the  Extraction  of  the  Giihe  Root. 

271.  In  extracting  the  cube  root,  we  formed  the  three  parts  of 
which  each  successive  remainder  was  composed,  regard  being  had  to 
the  denomination  of  the  tens  and  units  in  those  remainders.  But, 
since  any  number,  when  cubed,  gives  as  many  periods  of  three  figures 
each,  counting  from  the  right,  as  there  are  phices  of  figures  in  the 
given  number  (the  period  on  the  left,  however,  not  necessarily  contain- 
ing more  than  one  or  two  figures),  it  is  plain  that  the  first  two  periods 
on  the  left  of  the  given  number  give  the  first  two  figures  on  the  left  of 
the  root.  The  first  three  periods  on  the  left  give,  in  like  manner,  the 
first  three  figures  on  the  left  of  the  root,  and  so  on.  It  is  evident,  then, 
that  it  is  not  necessary  to  form  the  three  parts  of  which  the  successive 
remainders  are  composed;  we  have  only,  when  the  first  figure  of  the  root 
has  been  found,  to  cube  it,  and  subtract  the  cube  from  the  left  hand 
period.  "When  the  second  figure  of  the  root  has  been  found  by  using 
the  approximate  divisor,  3a'^,  we  have  only  to  cube  the  two  figures 
found,  and  subtract  the  cube  from  the  two  left  hand  periods,  and  bring 
down  the  next  period,  and  so  on.  If  the  cube  of  the  first  two  figures 
of  the  root  exceed  the  first  two  periods  on  the  left,  the  second  figure  of 
the  root  must  be  diminished  until  the  cube  of  the  two  figures  is  equal 
to,  or  less  than  the  two  periods  on  the  left.  The  successive  figures  of 
the  root  can  only  then  be  ascertained  by  trial.  We  can  always  tell,  by 
cubing  the  two,  three,  or  four  figures  of  the  root  found,  when  the  last 
figure  is  too  great ;  but  some  test  is  necessary  to  point  out  when  we 
have  diminished  the  last  figvire  too  much.  That  test  depends  upon  the 
principle,  that  the  difi"ercnce  between  two  consecutive  numbers  is  equal 
to  three  times  the  square  of  the  smaller  number,  plus  three  times  the 
smaller  number,  plus  unity. 

Let  a  be  the  smaller  number,  then  the   consecutive  number  next 


EXTRACTION    OF    ROOTS.  205 

above  it  will  be  a+1.  And  (a  +  lf—  {af  =  a^SaHSa  +  l  —  a» 
=  3a^  +  3a  +  1,  as  enunciated.  If,  tben,  after  subtracting  the  cube 
of  tlie  figures  found  from  the  periods  on  the  left,  the  remainder  is  ex- 
actly equal  to  three  times  the  square  of  the  root  found,  plus  three  times 
this  root,  plus  unity,  the  last  figure  of  the  root  can  be  increased  exactly 
by  unity.  If  the  remainder  is  greater  than  this  quantity,  the  last 
figure  can  be  increased  by  unity,  or  something  more  than  unity. 

The  following  examples  will  illustrate  the  foregoing  process.  Re- 
quired the  cube  root  of  531441. 

531441  I  «  +  i 
Sa'  =  19200  512  |     S'l 

19  441 
531  441  =  (a  +  If. 

After  finding  the  tens  of  the  root,  the  approximate  divisor  entered 
once  in  the  remainder,  the  quotient  was  set  on  the  right  of  the  root 
found,  and  the  two  figures  of  the  root  (81),  when  cubed,  gave  the 
original  number.  Hence,  the  number  was  a  perfect  cube,  and  81  its 
exact  root. 

Take,  as  a  second  example,  081472. 

G81472  \  a  +  b 
Sa'  =  19200  512  |     8  8 

169  472 

681  472  =  (a  +  by 

The  approximate  divisor  gives  9  for  the  second  figure  of  the  root ; 
but  89,  when  cubed,  exceeds  the  given  number,  681472.  The  last 
figure,  must,  therefore,  be  diminished,  if  we  make  this  figure,  7,  and 
cube  87,  the  cube  when  subtracted  from  681472,  leaves  a  remainder 
exactly  equal  to  three  times  the  square  of  87,  plus  three  times  87,  plus 
unity.  The  last  figure,  then,  can  be  increased  exactly  by  unity,  and 
88  is  the  true  root. 

272.  The  formula  for  the  difl'erence  between  consecutive  cubes, 
enables  us  to  pass  from  the  cube  of  one  number  to  the  cube  of  the 
number  next  above  it  without  actually  cubing  the  higher  number. 
Thus,  since,  the  cube  of  10  is  1000,  the  cube  of  11  must  be  1000  + 
3  (10/  +  3  (10) +1  =  1000  -f  300  +30  -fl  =  1331.  In  like  man- 
ner, since  (64)^  =  262144;  then  (65)="  =  262144  +  3  (64)^  +  3  (64) 
+  1  =  262144  +  12288  +  192  +  1  =  274625.  When  the  numbers 
18 


206  FORMATION    OF    THE    TOWERS    AND 

nuder  consideration  are  very  large,  the  formula  will  save  a  great  deal 
t)f  trouble  in  deducing  the  cube  of  the  higher  number. 

The  second  process  that  we  have  given  for  extracting  the  cube  root 
is  usually  followed,  and  when  the  given  number  contains  but  two 
periods,  is  shorter  than  the  first  process ;  but  in  every  other  case  is 
lonii-er. 


EXTRACTION  OF  THE  SQUARE  ROOT  OF  ALGEBRAIC 
QUANTITIES. 

273.  We  will  first  show  how  to  extract  the  root  of  monomials. 

Since  all  rules  for  the  extraction  of  roots  must  be  founded  iipon  the 
rules  for  the  formation  of  powers,  we  must  first  examine,  in  the  present 
ease,  how  the  power  of  a  monomial  is  formed.  Let  P  represent  the 
numerical  coefficient  of  any  monomial,  and  a"  the  literal  part  of  the 
monomial.  Then,  the  monomial  will  be  Pa"  j  now,  to  square  Pa"',  is 
to  multiply  it  once  by  itself.  And  Pa"  X  Pa™  =  P^a^".  Hence,  to 
square  a  -monomial,  we  have  only  to  square  the  coefficient,  and  to 
double  the  exponent  of  each  literal  factor.  The  square  of  2o-c  is,  by 
the  rvile,  4a*c^.  The  square  of  ba~^chl~^  is,  by  the  rule,  2ba~'^c'*d~*. 
The  square  root  of  4aV  is  then,  of  course,  2a^c;  but  —  2a^c  will,  when 
multiplied  by  itself,  give  4aV.  Hence,  the  root  of  4a''c^  may  be  either 
+  2a^c,  or  —  2a^c.  So,  in  like  manner,  the  root  of  25a~V<:?"''  may 
be  either  +  5a~'c^(?~^,  or  —  ba~^c^d~'^. 

RULE. 

I.  Extract  the  square  of  tlie  coefficient,  and  divide  tJie  exponent  of 
each  literal  factor  hy  2. 

II.  Write,  after  the  root  of  the  coefficient,  each  literal  factor,  tvith 
its  neio  exponent. 

III.  Affect  the  tchole  result  with  the  douUe  sign,  ±. 


EXAMPLES. 

1.  Extract  the  square  root  of  16a^c-^<f-".  Ans.  =b  Aa%-H-' 

2.  Extract  the  square  root  of  16a^"'i''"c^p.  Ans.  ±  4a"Z*''°cF 

3.  Extract  the  square  root  of  \AAan>'\'\  Ans.  it  12a''&V 

4.  Extract  the  square  root  of  3Ga'^?>'V°(^^l  ^^^    _t-  Qa^i\^°d'' 


EXTRACTION     OF    ROOTS.  207 

5.  Extract  the  square  root  of  a'^h'^c-'^^.  Ans.  ±  a^^h'rc-^^. 

6.  Extract  the  square  root  of  256 


7.  Extract  the  square  root  of 


a^b'c^'  Ans.   ±  a-'h-^c-^. 

25 

8.  Extract  the  square  root  of    .  ,  >  ,        _,    ^     i  -.  _i 

9.  Extract  the  square  root  of  ——-.  .         ,    =  i 

^  2o  Ans.  it  o~^xyz. 

10.  Extract  the  square  root  of  P-^M''-N-"=.       Ans.  ±  ^»3I-^N-^ 

11.  Extract  the  square  root  of    ,   ,.,    ,.  ,        _,    ^^  _„7_i„  _i 

^  «>i*''r         Ans.  ±  20a  "/^^"c-'. 

12.  Extract  the  sciuare  root  of  — j7r?r~  «         j    /■.-,^\■^-^  mUn 

^  400  .'Ins.   it  (20)  'a'"i'/v. 

13.  Extract  the  square  root  of  o-V>-^*c-'^^       Ans.  zt  n-7>-'v--^'-. 

It  is  plain  that  a  monomial  will  not  be  a  perfect  power  when  its  co- 
efficient is  not  a  perfect  square,  and  when  every  exponent  is  not  some 
multiple  of  2.  But  wlien  the  exponents  of  the  literal  factors  are  not 
exactly  divisible  by  the  index  of  the  root,  the  division  can  be  indicated. 

Tlius,  v/4x  =  =t  2.r-,  for,  by  the  rules  for  multiplication,  (db  2.r-) 

(=h  2x~)  =  ix.     Hence,  =t  2a:-  is,  truly,  the  square  root  of  -ix.     So, 

also,  VTiV  =  it  oH^c^,  because  (it  ah^c^)  (=t  ahh^)  =  al^c'. 
The  square  roots,  then,  of  all  algebraic  quantities  may  be  truly  ex- 
pressed whenever  their  coefficients  are  perfect  squares. 


SQUARE  ROOT  OF  POLYNOMIALS. 

274.  A  trinomial  is  the  least  polynomial  th^t  is  a  perfect  square. 
It  would  be  a  mistake  to  suppose  that  the  square  root  of  a^  +  Ir  is 
a  +  h,  for  (a  -f  hf  =  a^  +  2ah  +  l\  The  term,  lab,  enters  into 
the  square  of  (a  -f  h),  and  is  not  found  in  a'  -\-  1/.  Any  polynomial, 
to  be  a  perfect  square,  must  be  susceptible  of  resolution  into  two  equal 
factors ;  and  we  know  that  when  these  factors  are  multiplied  together  to 
reproduce  the  polynomial,  the  two  extreme  terms  of  the  product  (if  the 


208  F  O  11  M  A  T  I  O  N     OF     T  II  E     P  0  W  E  R  S     A  N  D 

factors  have  been  arranged  with  reference  to  a  certain  letter)  are  irredu- 
cible with  the  other  terms.  Hence,  the  square  root  of  these  extreme 
terms  must  be  terms  of  the  whole  root.  The  extreme  terms  of  s/a^-]-l/ 
must  then  be  a  and  h,  but  the  intermediate  terras  can  only  be  found 
approximatively. 

Any  binomial,  as  (a'  +  s),  when  squared,  will  give  a  trinomial,  a'^  + 
2a's  +  s^.  Conversely,  if  we  have  any  trinomial  that  is  a  perfect 
square,  its  root  must  be  a  binomial. 

Let  it  be  required  to  extract  the  square  root  of  ISah  +  81o^  +  7/. 
If  this  trinomial  be  arranged  with  reference  to  the  highest  power  of 
one  of  its  letters,  as  a,  and  the  square  root  of  the  first  term  be  taken, 
we  know  that  we  have  certainly  one  term  of  the  root  required,  Be- 
cause, from  what  has  been  said,  we  know  that  the  first  term  of  the 
arranged  trinomial  is  the  product  of  the  first  terms  of  the  two  equal 
factors  into  which  the  trinomial  can  be  resolved.  In  other  words,  it  is 
the  square  of  the  first  term  of  the  required  root.  The  arranged  poly- 
nomial is  Sla'^  +  ISab 
81a-  '+  ISah  -f  h'  I  9a +  b  +  h'.      We  begin  by 

Sla^  =  a'^  \lSa  +  b  =  2a's  +  s      extracting  the  root  of 

2a's  -f  s^  =  ISab  +  6^  the  first  term,  and  set 

18«6  +  b^  this  root  in  the  same 

horizontal  line  with  the 
polynomial,  and  on  its  right  separated  by  a  vertical  bar.  We  subtract 
from  the  given  quantity  the  square  of  the  first  term  of  the  root ;  there 
remain,  then,  only  the  two  terms,  18a?>  +  b^.  Now,  we  know  that 
the  first  term  of  this  remainder,  ISab,  is  the  double  product  of  the  first 
term  of  the  root  by  the  unknown  second  term.  If,  then,  we  divide 
18aZ»  by  2  (9a),  or  18a,  the  quotient,  b,  must  be  the  second  term  of 
the  root.  The  root  is  then  completely  known.  The  result  can  be 
verified  by  squaring  the  root  9a  -\-  b ;  or,  since  the  remainder  ISob 
+  b^  con-esponds  to  2a's  +  s'^,  and  since  18a  corresponds  to  2a',  if  we 
write  18a  below  the  root,  and  b,  which  corresponds  to  s,  on  its  right, 
and  then  multiply  18a  +  &  by  Z>,  we  must  evidently  form  the  remain- 
der, 2a's  -f  si 

275.  Since,  when  a  trinomial  is  a  perfect  square,  its  extreme  terms 
must  be  perfect  squares,  and  its  mean  term  must  be  the  double  product 
of  the  roots  of  these  terms,  we  can  tell  in  a  moment  when  a  trinomial 
is  a  perfect  square ;  the  mean  term  of  the  aiTanged  trinomial  must  always 
be  the  double  product  of  the  roots  of  the  extreme  terms.     Let  us  apply 


EXTEACTION     OF    ROOTS.  209 

tMs  simple  test  to  some  expressions.     4a^  +  4ma  +  rr^  is  a  perfect 
square,  and  its  root  2a  +  m  ;  a  +  2  V  a6  +  6  is  a  perfect  square,  since 

the  test  is  satisfied,  and  the  root,  -^ a  +  s/h  -,   .r^  -\-2.x-y~ -\-i/  is  a  per- 

3  3 

feet  square,  and  its  root,  x~  +  y~. 

The  root,  however,  will  only  be  commensurable  when  the  extreme 
terms  of  the  arranged  trinomial  are  rational.  We  have,  from  the  fore- 
going, a  simple  rule  for  the  extraction  of  the  root  of  a  trinomial  that  is 
a  perfect  square. 


Extract  the  root  of  the  extreme  terms,  and  connect  their  roots  together 
1)1/  the  sign  of  the  mean  term. 

Thus,  the  square  root  of  m^  —  2mn  +  n^  is  m  —  n;  this  is  obvious, 
since  {in  —  iif  =  m^  —  2mn  +  n^ ;  or  it  may  be  seen  by  going  through 
the  steps  of  the  process  described. 


EXAMPLES. 

1.  Kequircd  v/49a^  +  Warn  -f-  m^.  Ans.  la  +  m, 

2.  Required  v/49a''^  —  lAam -\- m^.  Ans.  la  —  vi. 

3.  Required  v/49a^  +  28am  +  4ml  Ans.  In  +  2m. 

4.  Required  y/^Qa^  —  28am  +  4m''.  Ans.  la  —  2m. 

5.  Required  V4m  +  IQ^Wn  +  16».  Ans.  2  n/w  +  4 v/TT. 

6.  Required  74m —  IQs/mTi  +  16rt7  Ans.  2-/m — 4%/?^ 

7.  Required  -/m^  —  14am  +  49al  A?is.  m  —  7a. 

8.  Required  V4hi2  —  28am  -f  49a''.  Ans.  2m  —  7a. 


Bema7-7:s. 

Examples  2  and  4,  in  connection  with  7  and  8,  show,  that  when  the 
mean  term  of  the  trinomial  is  negative  there  will  be  two  distinct  roots, 
according  to  the  arrangement  of  the  terms.  The  reason  of  this  is 
plain.  It  is  evident,  moreover,  that  when  either  of  the  extreme  terms 
has  the  negative  sign,  or  when  both  are  negative,  the  root  will  be 
imaginary. 

18*  o 


210  FORMATION    OF    TUE    POWERS    AND 

The  rule  for  the  extraction  of  the  square  root  of  a  trinomial  has  an 
important  application  in  the  solution  of  complete  equations  of  the  second 
degree,  and  ought,  therefore,  to  be  remembered. 

270.  If  the  given  numher,  instead  ofheing  a  trinomial,  have  a  tri- 
nomial root. 

Let  a  +  m  -\-  n  represent  this  root.  Then,  by  representing  a  +  m 
by  p,  and  the  given  polynomial  by  N,  we  have  N  =  (^  +  n)^  = 
jp2  _|_  2pn  +  n^  =  a^  +  2am  +  m^  +  2  (a  +  m)  «  +  n^.  We  see 
that  the  first  three  terms  is  a  perfect  square,  and  the  root  may  be 
found  by  the  rule  for  the  extraction  of  the  root  of  a  trinomial ;  then, 
when  we  have  taken  the  square  of  the  root,  a^  +  2am  +  ??i^,  from  the 
given  polynomial,  there  will  remain  2  (a.  +  m')  n  -f  n"^.  The  first  term 
of  this  remainder,  2an,  divided  by  2a,  the  double  of  the  first  term  of 
the  root  will  give  n,  the  third  term  of  the  root.  The  remainder, 
2  (a  +  m)rt  -f  n^,  may  be  put  under  the  form  (2a  +  2m  -f  71)%.  If, 
then,  the  first  two  terms  of  the  root  found  be  doubled,  and  the  third 
term  added  to  them,  and  their  sum  be  multiplied  by  the  third  term, 
the  product  thus  formed  will  be  equal  to  the  remainder.  The  process 
for  extracting  the  root  of  a  polynomial  which  is  the  square  of  a  trino- 
mial, is  precisely  like  that  for  extracting  the  root  of  a  polynomial  which 
is  the  square  of  a  binomial.  The  divisor,  to  get  any  term  of  the  root 
after  the  first,  is  twice  the  first  term  of  the  root ;  the  terms  of  the  root 
preceding  the  last  term  found  are  doubled,  the  last  term  added,  and 
the  product  of  the  whole,  by  the  last  term,  subtracted  from  the  succes- 
sive remainders. 

The  diifcrcnt  steps  can  be  best  exhibited  by  the  following  example  : 


a"  =  4:n' 


2m-f3m+5 


4n  +  3m 


2am+7n^-{-2  (^a+m)n-\-nr=127nn  +  dm^-\-i:nb-\-Qmb-\-h^    4?i-}-6m-f?» 

2am+m''=12«m-f9m^ 

2  (a  +  m)n-irn''=4:7ib+Qmb-\-b'^ 
(2a  +  2m+n)n=4:nb-^Qmb-\-b^ 

We  began  by  extracting  the  root  of  the  first  term,  the  root  found 
was  set  on  the  right,  and  its  square  subtracted  from  the  given  polyno- 
mial. We  had,  then,  taken  out  a^  of  the  formula  from  the  given  ex- 
pressions 5  we  next  doubled  the  root  found,  and  used  this  double  root 
(2rt  of  the  formula)  as  a  divisor  to  find  the  second  term.  This,  when 
found  (to  of  the  formula),  was  set  on  the  right  of  2??,  and  connected  by 


EXTRACTION     OF    ROOTS.  211 

its  appropriate  sign.  The  double  of  the  first  term,  and  the  second 
term,  written  directly  under  the  root,  were  next  multiplied  by  the 
second  term.  We  thus  formed  "lam  -\-  m^  of  the  formula,  which,  when 
subtracted  from  the  first  remainder,  left  -inh  +  6mh  +  l^,  correspond- 
ing to  2  (a  +  m)  71  +  ?i^  of  the  formula.  The  first  term  of  this  remain- 
der (2an  of  the  formula),  divided  by  twice  the  first  term  (2rt  of  the 
formula),  obviously,  gave  the  third  term  (?()  of  the  formula.  The  first 
two  terms  of  the  root  were  next  doubled,  and  the  third  term  added  to 
their  sum.  Having  thus  formed  2a  +  2m  +  m  of  the  formula,  wc 
multiplied  this  sum,  4?i  +  Qm,-\-h,  by  the  third  term,  h,  (or  n  of  the 
formula).  We  thus  plainly  formed  the  three  parts  of  which  the  re- 
mainder was  composed;  and  the  product  being  exactly  equal  to  the 
last  remainder,  the  polynomial  had  an  exact  root,  2«  +  3m  -|-  h. 

The  foregoing  reasoning  can  be  readily  extended  to  a  polynomial 
whose  root  contains  four  terms.  For,  let  N  represent  the  polynomial,  I, 
the  sum  of  the  first  three  terms  of  the  root,  and  n  the  last  term  of  root. 
Then,  '^  =  {I  -\- nf  =  F  +  2ln  -f  n^.  Suppose  the  first  three  terms 
to  be  a,  h  and  c;  then,  N  =  (a  -f  i  +  c)^  -f  2  (a  +  h  +  c)n  +  ii". 
Now,  the  first  three  terms  is  the  square  of  a  trinomial,  and  the  root  can, 
therefore,  be  extracted  precisely  as  in  the  foregoing  example.  After 
the  square  of  the  root  has  been  subtracted  from  the  given  polynomial 
there  will  remain  2  (a  +  b  +  c)  n+ii^.  The  divisor  to  find  n  is,  ob- 
viously, still  2a,  twice  the  first  term  of  the  root.  And,  since  th( 
remainder  can  be  placed  under  the  form  of  |  2  (a  -f-  S  4-  c)  -f-  »  |  n,  it 
is  plain  that,  if  the  last  term  be  added  to  twice  the  sum  of  the  first  three 
terms,  and  the  whole  sum  be  multiplied  by  the  last  term,  we  will  form 
the  three  parts  of  the  remainder.  The  process  for  extracting  the  root 
of  a  polynomial  which  is  the  square  of  four  terras,  is  then  identical  with 
that  for  exti-acting  the  root  of  a  polynomial  which  is  the  square  of  three 
terms,  and  that,  we  have  seen,  is  the  same  as  the  process  for  extracting 
the  root  of  a  polynomial  that  is  the  square  of  a  binomial.  Now,  if  the 
root  is  composed  of  five  terms,  after  the  square  of  the  first  four  terms 
has  been  subtracted  from  the  given  polynomial,  the  remainder  will  be 
found  to  be  the  double  product  of  the  four  terms  found  by  the  untnown 
fifth  term,  and  the  divisor  to  find  this  term  will  still  be  twice  the 
first  term  of  the  root.  And  so  the  reasoning  may  be  extended  to  a 
polynomial  whose  root  is  made  up  of  six,  seven,  or  any  number  of 
terms.  Hence,  for  extracting  the  root  of  any  polynomial,  we  have  the 
followins: 


212  FORMATION    OF    THE    POWERS    AND 


RULE. 

Arrange  the  polynomial  7vit7i  reference  to  one  of  its  letters.  Extract 
tlie  root  of  the  first  term,  and  set  the  root  in  the  same  horizontal  line 
icith  the  given  polynomial,  and  on  its  right,  separated  from  it  hy  a 
vertical  bar.  Suhtract  the  square  of  the  root  found  from  the  given 
polynomial,  and  hring  down  the  remaining  terms  for  the  first  remain- 
der. Write  double  the  first  term  of  the  root  found  immediately  be- 
neath the  place  of  the  root,  divide  the  first  term  of  the  first  remainder  by 
it,  and  write  the  quotient,  lohich  is  the  second  term  of  the  root,  on  the  right 
of  the  first  term  of  the  root,  and  also  beneath  and  on  the  right  of  double 
this  term.  Multiply  the  double  of  the  first  term,  and  the  second  term, 
itself  affected  with  its  propter  sign,  by  the  second  term,  and  subtract 
the  binomial  product  from  this  first  remainder,  and  bring  dotcn  the  re- 
maining terms  for  the  second  remainder.  Divide  the  first  term  of  the 
second  remainder  by  the  double  of  the  first  term  of  the  root,  and  the 
quotient,  affected  with  its  proper  sign,  will  be  the  third  term  of  the  root. 
Set  this  third  term  in  the  root,  and  also  in  another  horizontal  line  on 
the  right  of  double  the  sum  of  the  first  two  terms  of  the  root.  Midtiply 
the  three  terms  in  this  horizontal  line  by  the  third  term,  and  subtract 
their  product  from  the  second  remainder.  Continue  this  process  until 
the  final  remainder  is  zero,  the  root  will  then  be  exact;  or  continue 
until  the  letter,  according  to  which  the  polynomial  has  been  arranged, 
has  disappeared  from  one  of  the  remainders.  Tlien,  if  all  the  expo- 
nents of  that  letter  in  the  given  polynomial  are  positive,  we  conchide 
that  the  polynomial  is  not  a  perfect  square. 

Required  the  square  root  of  4x'^  +  Vlxy  +  4a;2  +  16x?  -f  %^  +  6^2 
+  24^?  +  %lz  +  16ZI 
The  polynomial  is  already  arranged. 

4z2  4- 12xy+4zz4-162:Z+97/2+6y2;+24?/Z4-8Zz+16Z2+22 
4x2 


1st  Rem.   =12zy+42z+16a;Z+V4-Gy3+24yZ+8/z+lGZ2-|-22 
]2zy  +V 


4X+32/ 


42:-f-6y+2 


4a;+6y+22+4i 


2cl  Rem.  =:4xz-\-16zl  -^Gi/z-\-24yl-^8lz-\-l&-^-\-z^ 

izz  +C.yz  2^ 

3d  Rem.  =16xZ  -^24yl-\-8lz-}-16l^ 

16x1  -\-24yl-{-Slz-\-lGP 


EXTRACTION    OF    ROOTS.  213 

277.  After  the  beginner  has  become  familiar  with  the  preceding 
principles,  it  will  not  always  be  necessary  to  go  through  the  process 
of  forming  the  successive  remainders.  The  successive  products,  distin- 
guished by  parentheses,  may  be  all  formed,  and  their  sum  taken  at 
once  from  the  given  polynomial,  as  in  the  following  example. 


TO*  —  6mn  +  Ore*  +  imp  —  2mq  —  \2np  +  6/17  -f  4p*  —  4p2  -f  5* 

(m'-i)  +  2(— 6j?m  +  9n»)  +  '(4mp  —  V2np  -j-  ip"^)  -f  *(—  •Zmg  +  (mq  —  ipq  +  g') 


7)1 — 3?i+2p— 7 


•2nj— 67i+2;> 


We  have  distinguished  the  successive  products  by  parentheses, 
affected  by  exponents  written  on  the  left. 

278.  It  is  obvious  that  the  successive  steps  required  by  the  general 
rule  for  extracting  the  square  root  of  any  polynomial,  amount  to  nothing 
more  than  subtracting  the  square  of  the  algebraic  sum  of  the  terms  of 
the  root,  as  they  are  found,  from  the  given  polynomial.  In  some  cases, 
then,  it  may  save  trouble  to  subtract  the  square  of  the  algebraic  sum 
of  the  first  two  terms  of  the  root  from  the  given  polynomial,  then  bring 
down  the  remaining  terms  and  find  the  third  term.  Next,  subtract 
the  square  of  the  algebraic  sum  of  the  first  three  terms  of  the  root  from 
the  given  polynomial,  and  continue  in  this  way  until  all  the  terras  of 
the  root  are  found. 


1.  Kequircd  v/x-  +  4^^  — Oxz  +  4ay  — 12?^  +  92^ 

Ans.  x-V^  —  ^z. 


2.  Required  V  x"  -^-i/  —  'ixY  +  4z''  +  \z^x'  —  \zhj\ 

Ans.  -x^  —  7f  -f  2z^ 


3.  Required  \  /  mi^  -f  ]/r  —  mn  +  -^  —  \xn  +  '\xm. 

Ans.  in  —  ^iU  -\-  Jx. 

4.  Required  \/a^ —  2«./'  +  x^  +  2am  —  'Ian  —  'Ixm  +  2xn.  -f  m^ 


2m?i  +  n^  Ans.  a  —  x  -\-  m  —  n. 


5.  Required  ^ *®  +  y**  +  txhn  +  2xy  —  'Ix^n  +  'li/hn  —  'Zijhi  + 


m^  —  2mn  +  n"^.  Ans.  x^  -\-  y^  -\-  m  —  n. 


214  FORMATION    OF    THE    POWERS    AND 


6.  Required  W  a'-  +  x^  +  2ax-{-am  +  a7i  +  xn  +  xm+  i7+x  +  ~9 


.  m        n 

Ans.   a  +  x  +  —  +  — 


\  Required  y/sLc^^  +  9a-?/+-^  +  9x2  +  9xm  +  ^  +  '!^'  +  i!  + 


4 


'"'  _L  '"  ^        n      ,    y    ,    2:    ,   m 

2"+  4"-  ^ns.  dx  +  ^+-  +  -. 

8.  Required  W  |-  +  ^  +  ^  -f  Qmi/  +  6mx  +  5^^  +  bzx  + 


36m^  +  202^  +  QOmz.  .         ^        V       n 

A71S.  _  +  ^  -f  6w  +  5z. 


9.  Required  v/^^  +  4x1/  +  4x'  +  1  +  4x  +  2y. 

Ans.  y  +  2x  +  1. 


10.  Required  n/^-''+  lOa-y  +  4a;'2/  +  12x2/''  +  ^V'- 

Alls  X"  4-  2x1/  -f  3y^ 


11.  Required  Vx*  -\-  2x^//  +  2x^y  +  2xY'  +  x^yt  +  ^V- 

Ans.  X?  -\-  xy  -{-  x?y. 

12.  Required  the  square  root  of  ^^  +  lOj^y  +  4ic^y +  12.r3/3+9y-|- 
2X'  +  ^x^y  +  10:rj/2  +  6/  +  x''  +  2.CTy  +  ?/^ 

Ans.  x''  +  2xj/  +  3^/2  +  a-  +  y. 

Remarlc. 

The  short  process,  indicated  in  Art.  277,  cannot  be  followed  in  the 
last  example. 


SQUARE  ROOT  OF  A  POLYNOMIAL  INVOLVING  NEGATIVE 
EXPONENTS. 

279.  The  principles  for  the  extraction  of  the  root  of  a  polynomial 
containing  negative  exponents  are  the  same  as  for  the  extraction  of  the 
square  root  of  a  polynomial,  all  of  whose  exponents  are  positive,  observ- 
ing, however,  in  the  arrangement  of  the  polynomial,  that  that  negative 
exponent  is  the  least  algebraically  which  is  the  greatest  numerically. 
When,  too,  the  arranged  letter  has  disappeared  from  any  remainder,  it 


EXTRACTION    OF    ROOTS 


215 


may  be  supplied  witli  a  zero  exponent,  provided,  that  the  exponent  of 
the  same  letter  is  negative  in  some  of  the  terms. 

Take,  as  an  example,  x-"^  +  2,x-^i/  +  x^if  +  2.xif  +2y  +  y^. 

Arranging  the  polynomial  with  reference  to  the  ascending  powers  of 
X,  and  proceeding  as  before,  we  have 


X-'  +  2x-'y  +  2xV  +  xy  +  2xf  +  x^ 

2x-'  +  y 

1st  Rem.  =:2x-'y  +  'Ix^y  +  xY  +  2x/  +  xhf 

2a;-'  +  2y  +  xy 

2x-\i/ 


+  xY 


2d  Rem. 


=  2x'>  +  2:?;y^  +  a;^ 
Ix'y  +  2x/  +  a;y 


The  first  term  of  the  second  remainder  is  2y ;  x,  the  letter  accord- 
ing to  which  the  polynomial  was  arranged,  docs  not  enter  into  this  re- 
mainder until  supplied.  It,  of  course,  must  be  introduced  with  a  zero 
exponent,  otherwise  the  expression  2y  would  be  altered  by  its  introduc- 
tion.    But,  since  x"  =  1,  then,  2x°y  =  2y. 

Take,  as  a  second  example,  x~*  4-  2x~^y  -f-  ?/-  -j-  2  +  2yx'  -i-  x*. 


X-*  +  2x-'y  -f  ^2  +  2  +  2yx'  +  x* 

^ 

1st  Rem.         =  2x-2y  +  /  -f  2  +  2yx'  -}-  x* 

'^''■rr'y  +  y' 

2d  Rem.  =  2x°  -f  2])x'  +  x* 

2x0  -f  2yx^  +  x* 


X-' 

+  y  +  ^ 

2x- 

-'  +  y 

2x 

-2  +  2y  +  x' 

The  first  term  of  the  second  remainder  is  2 ;  x,  affected  with  a  zero 
exponent,  was  introduced  into  that  terra. 

3.  Required  Vu;"^  -f  2x-'Y^  +  y^  +  2x-^  +  2xy-^  -f-  x\ 

Ans.  x~^  +  y-^  +  x. 

4.  Required  the  square  root  of  —7--\ ^ — [-"7--  +  1  +'^^—:^ — h-c"- 

4  o  y  o 

x~'^       i/~^ 


5.  Required  the  square  root  of  -j — | ^       '  .   i    .  "'  -^ 


+V  +  1  + 


0;-'  +  2a;-2-f^-|-4:c='-f-4. 


,-2       ,,-3 


Am.  _  -f-  ^  +  2  -f  xl 


216  FORMATION    OF    THE    POWERS    AND 

6.  Required  the  square  root  of  x^  —  2x^^  -}-  x*y^  +  2a;'2/~'  —  2x'  + 
2  —  2x-hj  +  2.r-='^-'  +  a"^  +  3/-I         A7is.  x"  —  x^i/  +  x-^  +^-'. 

7.  Required  the  square  root  of  x~^-{-2?/-\-x^i/'^-{-—'^ — l-a;~^^~'4-l. 


Ans.  X-*  H ^ h 


xy. 


INCOMMENSURABLE  POLYNOMIALS. 


280.  When  the  exponent  of  the  letter,  according  to  which  the  poly- 
nomial is  arranged,  is  positive  in  all  the  terms,  we  know  that  the  poly- 
nomial is  incommensurable  when  this  letter  is  not  found  in  any  of  the 
successive  remainders,  or  is  found  affected  with  a  lower  exponent  than 
it  appears  with  in  the  first  term  of  the  root.  But,  if  the  letter,  accord- 
ing to  which  the  arrangement  is  made,  appear  in  some  of  the  terms 
with  a  negative  exponent,  we  can  only  tell  that  the  given  polynomial 
is  not  commensurable  by  observing  that  the  operation  would  never  end. 


EXAMPLES 

1.  Required  the  square  root  of  a^  -f  h 

'  "^  2a        Sa'  ^  16^ 


,    ^'         ^'  I' 


2.  Required  the  square  root  of  1  —  x^. 

Ans. 

3.  Required  the  square  root  of  a;^  —  1 


X  X  X 

^ns.  l-^_--__&c. 


1  1  1  0 

2x        Qx''        Ib.c* 

4.  Required  the  square  root  of  2x^  -f-  2a;  +  2. 

13  ^ 

Ans.  xV2  -f  ^r  + =  +  &c.,  or  V^(x  +  1,  +  --  +  &G.) 

^/2       4x^2  ^         ~       Sx 

5.  Required  the  square  root  of  2x^  —  Gx  +  4. 

3  1  1 

Ans.  Xs/2 — __&c.,  or  y/2  (a-  — #— &c.") 

v/2       4x^/2  ^         -       8x  ^ 

6.  Required  the  square  root  of  to  -{-  n. 

Ans.    s/m  -j =. =  -\ ^^  —  &c. 

2s/m       StoVto       lGmV»i 

7.  Required  the  square  root  of  x~^  -f  2a;~^  -f-  4a~' 

Ans.  x-^  -{■  1  +  2x  -{■  &c. 


EXTRACTION     OF     ROOTS.  21' 


SQUARE  ROOT  OF  POLYNOMIALS  CONTAINING  TERMS 
AFFECTED  WITH  FRACTIONAL  EXPONENTS. 

281.  Since,  in  multiplication,  the  exponents  of  like  letters  are  added, 
whether  they  be  fractional,  or  entire,  it  is  evident  that  the  square  of 

m  m  m  Zm  2m  m 

rt""  =  a°  X  a"  is  a  ° .  Hence,  the  square  root  of  a^  is  a" .  Quanti- 
ties involving  fractional  exponents  may  then  be  operated  upon  in  the 
same  manner  as  quantities  containing  only  entire  exponents.  This  will 
be  shown  more  fully  under  the  head  of  fractional  exponents.  Assum- 
ing a  truth  which  scarcely  needs  a  demonstration,  we  will  give  a  few 
examples  of  polynomials  involving  fractional  exponents. 


EXAMPLES. 


1.  Required  ^/ x">  +  ,/0  —  2xy  —  2fz'  -f  2x'z^  -f 


Ans.  x"  — if'-\-  z". 


T.       •     ■,  \  /   ''        V        X  y  z        x^z         y  z 

Required  Vj  +  ^  +  -f-  +  -^-  +  -2-+V- 

J         i         ^ 
X-      y      z^ 

^"^-  Y  +  T+  2- 
Required  \/  -  +  -L  +  -^  ^  ^  +  — 


h^       2/^^ 


.        X         t/         z 
^ns.   _  -f-  4^  -f-  -. 
S         o         6 


4.  Required  \/x'^  +  y'^  +  2x'^y^'°  +  2^'0«^  -f  2a; 5 a^  +  a*. 

A71S.  x^  -{-  1/^^  -}■  d^. 

CUBE  ROOT  OF  POLYNOMIALS. 

282.  The  cube  of  a  monomial,  as  a,  is  a",  for  (ay  =  a  x  a  X  a  =za^. 
The  cube  of  a  binomial,  a  +  h,  is  a"  +  Sa^J  -f  3a&'  -f  b\ 
For,  by  actual  multiplication,  we  have  (a  +  ly  =  (a  +  li)  (a  +  h) 
(a  +  I)  =  a"  +  Sa'b  +  ^a¥  -f  V^.     Since,  then,  the  cube  of  a  mo- 
nomial is  a  monomial,  and  the  cube  of  a  binomial  a  polynomial  of  four 
19 


218  FORMATION     OF    THE    POWERS    AND 

terms,  it  follows  that  a  polynomial  of  four  terms  is  the  least  polynomial 
which  can  have  an  exact  cube  root. 

Knowing  the  third  power  of  a  binomial,  we  have  only  to  reverse  the 
process  to  obtain  the  cube  root  of  the  power.  We  see  that  the  first 
term  of  the  power  is  the  perfect  cube  of  a,  and  we  know,  from  the  man- 
ner in  which  a  product  is  formed,  that  that  term,  a?,  has  been  derived 
without  reduction  from  the  multiplication  of  the  three  factors  of  which 
the  power  is  composed.  If  then  we  extract  the  cube  root  of  o?,  we 
know  certainly  that  the  root,  a,  is  a  term  of  the  required  root.  Then  a 
will  be  one  of  the  extreme  terms  of  the  root;  and,  since  the  extreme 
terms  of  the  root  when  cubed  give  powers  that  are  irreducible  with  the 
other  terms,  it  follows  that  the  cube  of  a  or  a^  is  to  be  taken  from  the 
given  polynomial.  After  this  subtraction,  the  remainder  is  3a^6  + 
3a6^  +  1/,  and  it  is  plain  that  the  second  term  of  the  root  can  be  found 
by  dividing  the  first  term  of  the  arranged  remainder  by  3a^,  or  by  three 
times  the  square  of  the  first  term  of  the  root.  The  remainder,  3a^6  + 
3aZ>2  .^  ^3  ^^^  ^e  p^^  ^^^jgj,  tl^g  fQ^,j^  Qf  ^3^2  ^  3^  J  + 12)&.     If  then 

we  add  three  times  the  square  of  the  first  term  of  the  root,  three  times 
the  first  power  of  this  term  by  the  second  term  of  the  root,  and  the 
square  power  of  the  second  term  of  the  root,  together,  and  multiply  the 
sum  of  the  three  terms  by  the  last  term  of  the  root,  we  will  obviously 
form  the  three  parts  of  which  the  remainder  is  composed.  The  pro- 
cess, then,  for  extracting  the  cube  root,  is  analogous  to  that  for  extract- 
ing the  square  root  of  a  polynomial. 

a?  +  ^a'h  +  ^aW  +  V  \  a  +  h 


3a'  +  3a6  -f  6^ 


1st  Eem.  =  'da^h  +  'iay  +  W 


2d  Rem.  =0  0  0 

The  root  is  set  in  the  same  horizontal  line  with  the  given  polyno- 
mial, the  cube  of  the  first  term  is  taken  from  the  given  polynomial,  and 
the  first  term  of  the  arranged  remainder  is  divided  by  three  times  the 
square  of  the  first  term  of  the  root,  to  obtain  the  second  term  of  the 
root.  Finally,  immediately  beneath  the  root  is  written,  three  times 
the  square  of  the  first  term  of  the  root,  plus  three  times  the  product 
of  the  first  and  second  terms  of  the  root,  plus  the  square  of  the  second 
term  of  the  root,  and  the  product  of  the  sum  of  these  three  terms  by 
the  last  term  of  the  root  is  taken  from  the  first  remainder. 


EXTRACTION     OF     ROOTS.  219 

If  the  root  be  composed  of  three  terms,  a,  h  and  c,  we  may  repre- 
sent the  algebraic  sum  of  a  and  h  by  m.  Then,  let  P  be  the  poly- 
nomial whose  root  is  a  +  h  ■\-  c,  ytq  will  have  V  ^  (a  -\-  h  -{■  cf  = 
(m  +  cj  =  m^  -f-  ^m^c  -\-  Smc"  +  c"  =  (a  +  bf  +  3(a  +  bf  c  +  S 
(a  +  b)  e  f  c»  =  CT*  -f  3a^i  -f  ^ab^  +  i'  +  ^ah  +  Qabc  +  Z¥c  +  c^ 

The  first  four  terms  of  this  polynomial  are  the  same  as  in  the  last  ex- 
ample, and  therefore  their  root,  a  -f  i,  can  be  found  as  in  that  example. 
After  the  cube  of  the  sum  of  these  terms  has  been  taken  from  the  given 
polynomial,  the  remainder  may  be  placed  under  the  form  of  |  3  (a  +  by 
-f  3  (a  -f  6)  c-f  c^  I  c.  Then,  having  found  the  third  term  by  dividing 
the  first  term,  3aV,  of  the  arranged  remainder,  by  3a-,  it  is  plain  that  tha 
remainder  itself  can  be  formed  as  before,  by  multiplying  the  last  term 
found  into  three  times  the  square  of  the  sum  of  the  other  terms  of  the 
root,  plus  three  times  the  fii-st  power  of  the  sum  of  the  other  terms  into 
the  last  term,  plus  the  square  of  the  last  term. 

Now,  whatever  may  be  the  number  of  terms  in  the  root,  we  may  re- 
present the  algebraic  sum  of  all  of  them,  except  the  last,  by  s.  Sup- 
pose n  to  be  the  last  term.  Then  P  =  (s  -f  ?i)'  =  s'  +  Ss^n  +  Ssn^  +  ?i' : 
in  which  s  represents  the  algebraic  sum  of  any  number  of  terms.  If 
a  is  the  first  term  of  s,  then  the  first  term  of  the  development  of  3s^m, 
will  be  Sa^n.  Hence,  the  divisor  to  find  n  is  still  three  times  the 
square  of  the  first  term.  After  s'  has  been  subtracted  from  the  poly- 
nomial, the  remainder  can  be  put  under  the  form  of  (3s^  -|-  Ssn  +  n^)«. 
Hence,  whatever  the  number  of  terms  in  the  root,  the  successive  re- 
mainders can  be  formed  just  as  when  there  are  but  two  or  three  terms 
in  the  root. 

The  following  examples  will  illustrate  the  process. 


1st  R.=3x»+3a;*+9x*y+x=+18x=^+27xV+9-c'i'+27ii/^-|-  27^' 
3x5  ^sx*  +x'>  


(3x«  +  3x3  _^  a.»)^ 


(3(x^+^)»+3(xHa;)3y+9y»)3y 


Rem.  =  9x*y  +  18x=(/  -f  27xV  -|-  9x^y  +  27x3/«H-  27  j' 

9x«y  -I-  ISx^y  +  27xV  +  9x»y  +  27xy''  +  27y» 
Rem.  =00  00  00 


x'+  Qx'+  15x*+20x'+lox'+Qx+ 1 
x' 


1st  E.  —Qx^i-  15x'-f  20x'+  ldx%  Gx  +1 
Gx'+12x'+  8:c^ 


a;^-f2.c-f  1 


(3x^-f-6.cH4.r^)2.T 


(;6(x'+2xy+S(x'+2x)l+l)l  = 
(Sx'+  12x='-fl2xH3x^-h  6x+ 1)1 


2d  Rem.     =  Sx*+  12x^-f  15x-^+  6x  -j- 1 

3.r^+12.T='+15x^-f6x+l 

3d  Piem.     =  T»  0  0       0~0 


220  FORMATION     OP    THE    POWERS    AND 


RULE. 

Arrange  the  given  polynomial  with  reference  to  the  ascending  or  de- 
scending powers  of  one  of  its  letters.  Extract  the  ciile  root  of  the  term 
on  the  left,  and  set  the  root  on  the  same  horizontal  line  toith  the  given 
polynomial  on  the  right,  and  separated  hy  a  vertical  line.  Subtract 
the  cuhe  of  this  first  term  of  the  root  from  the  given  polynomial,  and 
hring  down  the  remaining  terms  for  the  first  remainder.  Write  three 
times  the  square  of  the  first  term  of  the  root  immediately  beneath  the 
place  of  the  root.  This  will  be  the  divisor  to  find  all  the  other 
terms  of  the  root.  Divide  the  first  term  of  the  remainder  by  the  divisor, 
and  set  the  quotient,  with  its  appropriate  sign,  on  the  right  of  the  first 
term  of  the  root,  for  the  second  term,  of  the  root.  Set,  on  the  right  of 
the  divisor,  three  times  the  product  of  the  first  and  second  terms  of  the 
root,  affected  with  its  proper  sign,  plus  the  square  of  the  second  term  of 
the  root.  Next,  midtiply  the  algebraic  sum  of  the  three  terms,  thus 
formed,  by  the  last  term  of  the  root,  and  subtract  the  product  from  the 
first  remainder.  Bring  down  the  remaining  terms  for  the  second  re- 
mainder, and  divide  the  first  term  by  the  divisor.  The  quotient  is  the 
third  term  of  the  root.  TaJce  three  times  the  square  of  the  sum  of  the 
first  two  terms,  and  add  to  this  three  times  the  product  of  the  algebraic 
sum  of  the  first  two  terms  by  the  third  term, phis  the  square  of  the  third 
term.  Multiply  the  algebraic  sum  of  the  whole  by  the  third  term,,  and. 
subtract  the  product  from  the  second  remainder.  Bring  doivn  the  re- 
maining terms  from  the  third  remainder.  Find  the  fourth  term  of 
the  root  as  before,  and  continue  the  process  iintil  all  the  terms  of  the  root 
are  found. 

Remarks. 

1st.  When  the  exponent  of  the  letter,  according  to  which  the  ar- 
rangement is  made,  is  positive  in  all  the  terms  of  the  given  polynomial, 
we  know  that  the  root  is  not  exact,  whenever  the  exponent  of  the  as- 
sumed letter  is  less  in  the  first  term  of  any  of  the  successive  arranged 
remainders  than  it  is  in  the  divisor. 

2d.  It  is  evident  that  the  steps  of  the  process,  according  to  the  rule, 
amount  to  nothing  more  than  subtracting  the  cube  of  the  algebraic  sum 
of  the  terms  of  the  root,  when  found,  from  the  given  polynomial.  It 
may,  therefore,  sometimes  save  trouble  to  proceed  thus.  Subtract  the  cube 
of  the  first  term  from  the  given  polynomial.     Find  the  second  term  of 


EXTRACTION    OF    ROOTS.  221 

the  root  according  to  the  rule.  Cube  the  algebraic  sum  of  the  two 
terms  of  the  root,  and  subtract  the  result  from  the  given  polynomial. 
Find  the  third  term,  and  subtract  the  cube  of  the  algebraic  sum  of  the 
first  three  terms  of  the  root  from  the  given  polynomial.  Proceed  in 
this  manner  until  we  get  a  zero  remainder,  or  until  it  becomes  evident 
that  the  given  polynomial  is  incommensurable. 

EXAMPLES. 

1.  Eequired  the  cube  root  of  x^  -f  3x«  +  3x^  +  7x«  +  12x*  +  Qx* 
+  12x='+12x2+ 8.  Aus.   x'^+x^^^. 

2.  Kequircd   the   cube   root   of   x^  +  6x'  +  21a;*  +  44a;'  +  63^-^ 
+  54x  +  27.  Ans.  x^  4. '2x  +  3. 

3.  Eequired  the  cube  root  of  8a;«  +  48x'  +  lG8x^  +  352ar'  +  504x2 
+  432:c  +  216.  Ans  2x'  +  4a;  +  6. 

4.  Required  the  cube  root  of  tt  +  x  +  ~yr  +"97"+  — /^  +  -^  +  1- 

,        x^       x       ^ 
Ans.-  +  j  +  \ 

r    -n       .     ,,,         ,  ,    ,.  125a;'       75.?;"'       15a;'       a;'' 

5.  llcquired  the  cube  root  of  ■     .  -  -\ -7-  4 1 . 

^  64     ^    64   ^    64    ^  64 

Sa;^'       la; 
An,.  ^+Y- 

p     T>        •     J  xu         1-          i.    ^  125a;'       7bx''      15.';*       a;' 
0.  Required  the  cube  root  of 1 1 1 . 

5a;'      la; 
Ans.-~  +  -. 

r,    T,       .     ,  ,,  ,  ^        ■x\    Sx'      3.r'       7a;«       Sx'       Sx* 

7.  Required  the  cube  root  of  —  +-— +  —-  +  —-  +  __  -| 

8         8         8         8  2  4 

,  3a;'      3x^  ,  ^  .        ^'     ^      , 

8.  Required   the    cube  root   of   Scv^  +  12a' J  —  12a^m  +  Qah'  — 
12amb  +  Qant"  —  3mb^  +  Sm'b  +  6'  —  m'  Ans.  2a  +  b^m. 

9.  Required  the  cube  root  of  a;' — 9x^^  —  Sx^  +  27a;/ +  18a;y 
+  3x  — 27/— 272/^  — 9y  — 1.  ^«s.  a;  — 3^  — 1. 

10.  Required  the  cube  root  of  x^  +  3x*  +  3a;y  —  Sa;^  +  Sa;"  +  a;' 
+  6.xy  —  6a;'y  +  6a;V  +  SxY  —  QxY  —  Zxhj  +  3.r/  —  6x/  + 
3a;/  +  /  —  3/  +  3/  — /.  Ans.  x^  +  x  +  /  —y. 

19* 


222  FORMATION    OF    THE    POWERS    AND 

CUBE  ROOT  OF  INCOMMENSURABLE  POLYNOMIALS. 

283. — EXAMPLES. 


1 .  Required  the  cube  root  of  x^  +  3j;^  +  3.r  +  3 

Ans.  rr  +  ] 

2.  Required  the  cube  root  of  x^  +  Qx^  +  12a;  +  9. 


2 
Ans.  rr  +  1  +  g-^+,&c. 


Ans.  a^  +  2  +  ^^-^  +  ,&c. 
ox 


3.  Required  the  cube  root  of  x^  +  o^. 

Avic       nr    _L  . 

3^2      9a-.^' 

4.  Required  the  cube  root  of  x^  +  c,x^  +  4. 


a"        a" 


^ns.  ic  +  1 ,  &e. 

X 


5.  Required  the  cube  root  of  x  +  3.c^  +  6. 

6.  Required  the  cube  root  of  u^  +  Zar^  +  3x-i  +  2 


y/x} 


Ans.  .r~'  +  1  +  - — ;r-  +  ,  &c. 


7.  Required  the  cube  root  of  x~^  +  &x~'^  +  12.x~'  +  9 

3^ 


Ans.  X  >  +  2+o^+,&c. 


8.  Required  the  cube  root  of  x~^  +  3a;~^  +  3.t~^  +  2. 

Ans.  x~'^  _j_  1  --(-- 

9.  Required  the  cube  root  of  x''  +  9.r«  +  27.r'  +  26 

^72S.     .T^  -j.  3 

10.  Required  the  cube  root  of  x^  +  Scc^  +  3a;*  _}-  a; 


Ans.  x-2  j_  l.f-J_-f  &e. 


^72s.  a-3+3  — —  — ,&c. 


Ans.  x^  -\-  X  —  K J  &c. 


EXTRACTION     OF    ROOTS.  223 

CUBE    ROOT    OF    POLYNOMIALS   INVOLVING    FRACTIONAL 
EXPONENTS. 

284.  —  EXAMPLES. 

3  1 

1.  Kequired  the  cube  root  of  x~  +  3^:;  +  3x-  -f  1. 


7  5 

2.  Required  the  cube  root  of  x  +  Sz^  +  3x^  +  x 


3  9 

3.  Required  the  cube  root  of  .t«  +  3.c'  +  x-  +  3x- 


Ans.  a;-  +  1. 

1 

3 

1  1 

Ans.  x^^  +  x^. 
I 

ins.  x^  +  a:-. 


5_1  L3  3 

4.  Required  the  cube  root  of  .x'^  +  3x  ^   +  ox  ^   +  x'^. 

Ans.  x^  +  x^. 

5.  Required  the  cube  root  of  .r«  +  3x^  +  3a;*  +  3ar'  +  Gx^  +  Sx^ 
+  x^  +  Bx  +  8x2  4.  1.  ^„s.  0-2  +  x2  +  1. 

6.  Required  the  cube  root  of  8x«  +  2-kJ  +  24x''  +  48x'^  -f  24x=' 
+  8x2  4.  24x  +  24x2  ^  8.  Aiis.  2x^  +  2x2  +  2. 


1  4  ]j? 

7.  Required  the  cube  root  of  x2  +  3x^  +  3x  •*  +  x' 


Ans.  x^  +  X. 


11  17 

8.  Required  the  cube  root  of  x^  +  Sx*^  +  Sx^  -\-  x^  . 

1  J  J 
Ans.  x^  +  x'«. 

1  fi  i_4  4 

9.  Required  the  cube  root  of  x^  +  3x  ^   +  3x  "  +  x^, 

2  4 
Ans.  x3  -|-  x^. 

10.  Required  the  cube  root  of  x'^  +  3x^4^  +  3x'^^  +  1  +  3x^  +  3x2 
+  6x^4'^  +  3x*  +  3x^  +  xl  Ans.  x'  +  x4  +  1. 

11.  Required  the  cube  root  of  x~'^  +  3x~  ^   +  3x~2  -j.  1  _j-  S^c  »  _f- 
6x~'4    4-  Sx~~  +  3x-'  +  x~4  +  Sx~4.  Ans.  X-*  +  x~4  +  1. 


224  FORMATION     OF    THE    POWERS    AND 

12.  Required  the  cube  root  of  x"'^  +  3x~  s   +  3x~  §  +  x~^. 

Ans.  x~^  +  *~^- 

L3  8 

13.  Required  the  cube  root  of  cc^  —  3x  ^   -f  3x3  —  ^. 

Ans.  x^  —  x^ . 

14.  Required  the  cube  root  of  x^  +  3*^  +  Qx^  +  ^x'^  +  Qx^~ +^x^ 
+  lOx^  +  9«2  +  1x^^  +  6x  +  3a;^  +  1.        Ans.  x^ -^  x -^  x^  + 1. 

_3  _S 

15.  Required  the  cube  root  of  1  +  3x-'  +  6j3~^  +  3a;  ~  +  6     ~  + 
9.7;~3  +  lOx-^  +  9x-^  +  7a;~^  +  6x-^  +  3x~'^'  +  »-«. 

Ans.  1  +  X-'  +  x~3  +  a;-l 
IG.  Required  the  cube  root  of  27xH81x5+162x''  +  81«~  +  162a;^  + 
243a;^  +  270a;='  +  243a;H189x"^'  +  162x  +  Slx^  +  27. 

Ans.  3  (a;'^  +  a;  +  a;^  + 1). 


CUBE  ROOT  OF  POLYNOMIALS  CONTAINING  ONE  OR  MORE 
TERMS  AFFECTED  WITH  NEGATIVE  EXPONENTS. 

285.  The  process  is  the  same  as  for  polynomials,  all  of  whose  terms 
are  affected  with  positive  exponents;  observing,  in  the  arrangement, 
that  that  negative  exponent  is  the  least  algebraically  which  is  the  great- 
est numerically,  and  that  when  the  letter,  according  to  which  the 
arrangement  has  been  made,  has  disappeared  from  any  of  the  succes- 
sive remainders,  it  may  be  introduced  affected  with  a  zero  exponent. 


EXAMPLES. 

1.   Required  the  cube  root  of  x~^  +  3  -[-  3x^  -f-  x^. 

Ans.  x~^  -f  cc*. 
For,  by  the  rule 

x-^  +  3  +  3x«  +  x"^  I  x-"-  -f  a* 


(3x-*  +  3a;2  -f  x^'^x" 


1st  Rem.  =  3x°  -f  3.^^  +  x'^ 


2d  Rem.  =  0 


EXTRACTION    OF    ROOTS.  225 

This  polynomial  miglit  have  been  arranged  with  reference  to  the  de- 
scending powers  of  x. 

2.  Required  the  cube  root  of  :r~^  +  Sji-^  +  3  +  a;'- 

Alls.  cc~^  +  X. 

3.  Required  the  cube  root  of  x-^  +  3:e-^  +  3x-^  +  3x-2  +  6x-'  + 
4  +  3^  +  3^2  +  x^.  Am.  x-''  -\- x  ■\-  1. 

4.  Required  the  cube  root  of  x'  +  S.r^  +  ox  +  7  +  12x-'  +  6x-^  + 
12x-^  +  12x-^  +  8x-«.  ^«s.  a;  +  1  +  2^-^ 

5.  Required  the  cube  root  of  I  +  Gx"'  +  21.>:-^  +  44x-^+  63x-^  + 
54x-^  +  27x-«.  Am.  1  +  2x-'  +  Sx"''. 

6.  Required  the  cube  root  of  a"^  —  Ga-^i  +  12a-^Z<'  — Sa-«6'. 

^Hs.  a-'  — 2a-^6. 

7.  Required  the  cube  root  of  x^  +  Gx^  —  40  +  GGx-^  —  G4x-^. 

Ans.  X  -f  2  —  4x~'. 

8.  Required  the  cube  root  of  1  +  Gx"'  —  40x-^  +  OGx-^  —  G4x-«. 

Am.  1  +  2x-'  — 4x-2. 

9.  Required  the  cube  root  of  1  —  Gx"'  +  lox-^  —  20x-'  +  15x-'  + 
—  6x-5  +  x-«.  Am.  1  —  2x-'  +  x-^ 


QUANTITIES  AFFECTED  WITH   FRACTIONAL  EXPONENTS. 

28G.  Suppose  it  be  required  to  multiply  a"  into  itself  until  it  is  taken 

three  times  as  a  factor.     Then,  by  the  rules  for  multiplication,  «"  X  a" 

X  a    =  a'-.     This  result  is,  evidently,  the  cube  of  a".     Now,  o-jmul- 

.1        J         3  

tiplied  by  itself  once,  gives  a'  X  a'  =  a~  =  a ;  and  ^/ a,  multiplied  by 

itself  once,  gives  -J a  X  \/  a  ^  yj  a-  =  a.     So,  a',  and  \/a,  arc  cqui- 

valent  expressions.     The  expression,  a"-,  then,  which  indicates  the  cube 

of  a",  also  indicates  the  cube  of  sf  a.     The  numerator  of  the  fractional 
exponent  denotes  the  power  to  which  the  quantity  has  been  raised,  and 

the  denominator  the  degree  of  the  root  to  be  extracted.     Taking  a 

three  times  as  a  factor,  we  have  a   X  a   X  a    =  a^"*"^"*"^  :=  a.       Now, 
P 


226  FORMATION     OF    THE    POWERS    AND 

the  ^/ a,  raised  to  the  third  power,  will  also  plainly  be  a,  because, 
from  the  definition  of  Involution  and  Evolution,  it  is  evident  that 
the  latter  undoes  what  the  former  has  done.  The  root  of  any  quantity 
raised  to  a  power  indicated  by  the  index  of  the  root,  must  then  be  the 

quantity  itself.     Hence,  (  ^  a)*  :=  a  ;  then,  a=  ^  a ;  since  we  have 

seen  that  both  a  ,  and  V  a,  when  cubed,  give  a.     In  general,  an  = 

^a".     Because  aa,  taken  as  a  factor  n  times,  gives  d^Xa^Xa^  = 

a"  "  n  =  a",  and  V«"j  raised  to  the  n^^  power,  is  also  a™.  The 
fractional  index  may  be  regarded  as  denoting  a  poioer  of  a  root.  The 
denominator  expresses  the  root,  and  the  numerator  the  power.  The 
denominator  shows  into  how  many  equal  factors,  or  roots,  the  given 
quantity  is  divided,  and  the  numerator  shows  how  many  of  these  fac- 
tors have  been  taken.  The  fractional  index  may  be  considered  to  de- 
note tJie  root  0/  a  power,  as  well  as  the  power  of  a  root.  Thus,  a^  may 
either  denote  that  the  third  power  of  the  *J  a  has  been  formed,  or  that 
the  square  root  of  the  third  power  has  to  be  taken.  Taking  the  former 
view,  it  indicates  an  executed  operation ;  taking  the  latter  view,  an 
unexecuted  operation. 

In  general,  a"  may  be  read  the  m^'^  power  of  the  n^^  root  of  a,  or  the 
«."'  root  of  the  m}^  power  of  a.  It  is  more  convenient  to  read  the  ex- 
pression thus,  a  to  the  m  divided  by  n,  power ;  or  a,  raised  to  a  power 

denoted  by  the  quotient  of  m,  divided  by  n.     So,  ai  is  read,  a  to  the 

P 

—  power. 

9 

There  are  two  consequences  of  notation  by  means  of  fractional  expo- 
nents that  deserve  consideration.  1st.  Any  multiple  of  the  numerator 
of  a  fractional  exponent  may  be  taken,  provided  an  equal  multiple  of 

the  denominator  is  also  taken.    Thus,  az=a^  is  also  n=a.^^=a^=a"',  &c. 

So,  also,  ai  =  a'l.  The  reason  of  this  is  plain;  the  increase  of  the 
denominator  makes  the  equal  roots  or  factors  of  the  given  quantity,  a, 
smaller;  but  the  numerator  being  increased  just  as  much  as  is  the 
denominator,  the  number  of  these  smaller  equal  roots  will  be  increased 
enough  to  make  up  for  their  diminution  in  magnitude.  2d.  Since  the 
fractional  exponent  may  be  exchanged  for  any  other  of  equal  value,  it 

may  be  expressed  in  decimals.      Thus,  o^  =  a"*^  ~  a-^.     So,  also,  «•'» 


EXTRACTION    OF    ROOTS.  22 1 

=   a.037037+_ 

These  decimal  indices  are  called  logarithms.     The  manner  of  calcu- 
lating them,  and  their  use,  will  be  shown  hereafter. 


MULTIPLICATION  OF  QUANTITIES  AFFECTED  AYITII  ENTIRE 
OR  FRACTIONAL  EXPONENTS.  —  MONOMIALS. 

287.  Quantitities  with  fractional  exponents  must  plainly  be  multi- 
plied in  the  same  way  as  quantities  affected  with  entire  exponents. 
That  is,  the  exponents  of  the  same  letter  or  letters  in  the  multiplicand 
and  multiplier  must  be  added,  and  after  the  common  letter  or  letters 
must  be  written  tho.ee  not  common,  with  their  primitive  exponents. 

Let  it  be  required  to  multiply  a^  by  o^.     Then,  a-  is  to  be  repeated 

a^,  or  a-  times.     Now,  to  repeat  a-,  a-  times,  we  have  only  to  add 
the  exponents  of  multiplicand  and  multiplier,  for  the  first  exponent 

1  3 

denotes  that  a^  has  been  taken  three  times  to  produce  the  product,  a-\ 
and  the  second  exponent  denotes  that  a^  has  been  taien  four  times  in 

4  1 

the  product,  a^.     Hence,  a^  must  enter  seven  times  in  the  result  of 

3  4  .7 

the  multiplication  of  a-  by  a-,  and  that  result  must  be  written  a^.  To 

-1.1  i  1 

multiply  a-  by  h,  is  to  repeat  a-,  h  times.     Hence,  a-xh  =  a^b. 

RULE. 

Multiply  the  coefficients  together  for  the  coefficient  of  the  product. 
Reduce  the  exponents  of  the  same  letters  in  multiplicand  and  multiplier 
to  the  same  denominator,  add  their  numerators,  and  set  their  sum  over 
the  common  denominator.  Annex  these  letters  to  the  new  coefficient, 
and  lorite  after  them  all  the  letters  ichich  are  not  common  to  the  multi- 
plicand and  midfiplier.  '-. 

EXAMPLES. 

112       3  3        4 

1.  Multiply  a^6°"  by  a°Z>^c.  Ans.  n^l>''c. 

i    J    1  7     13    9 

„    ,,  .  .  ,     a^McS  ^     aWc+'  ,        a^b  4  c^ 

2.  Multiply  -—^—  by  ^— .  Ans.   ^ . 


228  FORMATION    OF    THE    POWERS    AND 

3.  Multiply by .  Ans. 


4.  Multiply     ^    ^   J  by  a~x^i/^z^. 
x-i/z" 


5.  Multqily  a  "6   "c  p  by  a"i"'cp . 


-4/is.  aa-^z. 

Am.  1. 

s-2r    q-3p 

^ws.  a  2^  6  3q  ^ 

1  J^ 

Ans.  —  56x^3/4% '^0. 

^ns. 

2no+l     3d+1   4p-|-l 

60a   2   V^G  *  . 

6.   Multiply  a    ^b    "Jc    -  by  a'i^c^ 

3      4      4 

I.  Multiply  2a   •'b  "c    "  by — - — . 


8.  Multiply  Sx'^y^z'"  by  —  7x^bK^. 

9.  Multiply  12a"'i-cP  by  5a'' 6  c*. 

_^   _1  __L  "    _L  -P 

10.  Multiply  4a  "  i   =c   i    bya°6»ci.  J-«s.  4. 


DIVISION  OF  QUANTITIES  AFFECTED  WITH  FRACTIONAL 
AND  ENTIRE  EXPONENTS.  —  MONOMIALS. 

288.  Since  tbe  exponents  of  tbe  like  letters  are  added  in  multiplica- 
tion, they  must  be  subtracted  in  division.  For,  the  object  of  division 
is  to  find  a  quantity  called  the  quotient,  which,  when  multiplied  by 
the  divisor,  will  give  the  dividend.     Let  it  be  required  to  divide  a^  by 

a'  ;  then,  we  must  find  a  quantity,  which,  when  multiplied  by  a"  will 


=  a    —  a 


give  a^.     This  quantity  is  plainly  a'~,  for  a    X  a~  =  a     ^ 

3 

But  the  quantity  found,  a~ ,  has  resulted  from  the  subtraction  of  the 

exponent  of  a"'  from  2  =  4,  the  exponent  of  a^.     So,  the  result  of  the 

I  1  "        i_  1  J._  L        '  i 

division  of  ci^  by  a°^  must  be  a"'  ",  since  a"  '•Xa»=a'".  When 
the  denominators  of  the  fractional  exponents  are  diff"erent,  the  subtrac- 
tion can  only  be  indicated,  not  performed,  until  they  are  reduced  to 
the  same  index.  If  the  exponents  of  the  same  letter  in  the  divisor 
exceed  that  of  the  dividend,  the  exponent  of  that  letter  in  the  quotient 


EXTRACTION    OF    ROOTS.  229 

will  be  negative.  So,  also,  if  there  are  any  letters  in  the  divisor  not 
coiumon  to  the  dividend,  they  must  appear  in  the  quotient,  with  their 
primitive  exponents  taken  with  a  contrary  sign ;  else,  when  the  quo- 
tient and  divisor  are  multiplied  together,  these  letters  would  enter  in 
the  product. 

RULE. 

Divide  the  coefficient  of  the  dividend  hy  that  of  the  divisor,  the  result 
icill  he  the  coefficient  of  the  quotient.  Write  after  it  all  letters  common 
to  dividend  and  divisor,  affected  with  exponents  equal  to  the  difference 
of  their  exponents  in  the  dividend  and  divisor ;  and,  also,  cdl  the 
letters  common  to  the  dividend  only,  with  their  primitive  exponents; 
and  all  common  to  the  divisor  only,  icith  their  primitive  exponents 
taken  with  a  contrary  sitjn. 


3     1  .1    i     }  3       _    1 

1.  Divide  4a2c-(?'^  by  2a"c'cZ.  Ans.  2a~cd   ^^. 

11}  1  °^'"  '-2«  t 

2.  Divide  VZa^b' c  d^  by  Ga- b"".  Ans.  2a  »' b'^c  d^. 


3.  Divide  7x"'?/"zp  hj  x'"y''z~p.  Ans.  Iz  p  . 

_j  _j          J  I   1  _i 

4.  Divide  72  "rj      hjz'y  x  .  Ans.  Tar'^-'x     . 

5.  Divide  24a'*i^c^  by  242='a~''i~^c"^.  Ans.  z'^ahc. 

h  +  -}  18 

6.  Divide  l^x" y"" z    by  — j — j — -. 

a;~V  2~*  Ans.  1. 

7.  Divide  oa-^h    '  c"  by  h^'vh    ^d  "■ .  Ans.  9i '  c^J" . 

8.  Divide  80m"r'"i^c^  by  5m^j"^;r".  Ans.  IQm'Hc^x'^. 

_1      1     _i_  (m+n         (l+n)         (I+p) 

9.  Divide  360e  -/"^   p  by  1 80^7^.     Ans.  2e~~^f     ^g~'^. 

10.  Divide4a"c^&''  by  a-c^~^e~*.  .4ns.   ch^d^^e\ 
20 


230  FORMATION     OF    THE    POWERS    AND 


RAISING  TO  POWERS  QUANTITIES  AFFECTED  WITH  FRAC- 
TIONAL AND  ENTIRE   EXPONENTS.  —  MONOMIALS. 

289.  Since,  to  raise  a  quantity  to  a  power  is  only  to  multiply  it  by 
itself  a  certain  number  of  times,  the  rule  for  the  involution  of  quantities 
affected  with  any  exponents  whatever  is  deduced  directly  from  the  rule 
of  multiplication. 

RULE. 

Raise  the  coefficient  to  the  required  power,  and  write  after  it  all  the 
literal  factors  affected  with  exponents  equal  to  the  product  of  their 
primitive  exponents  hy  the  exponent  of  the  power. 

Thus,  to  raise  a/  to  the  third  power,  is  to  take  a     three  times  as  a 

''  \  i  2"  2+^+^  ^ 

factor.  Then,  (a")'' =  a  Xa''y.a  =a  "  =a-^.  The  exponent 
of  the  result  is  the  product  arising  from  multiplying  the  exponent  of 

1  m 

the  quantity  by  3,  the  exponent  of  the  power.     So,  (Mx"  y=(M.yx° , 

I  J_  L   1 

because,  (Ma;")"  =  M  X  M  X  Ma;"' +'""'''"  +  &c.  If  the  given  quan- 
tity be  a  fraction,  it  is  raised  to  the  required  power  by  raising  the 
numerator  and  denominator,  separately,  to  that  power :  because,  the 
power  of  a  fraction  is  nothing  more  than  the  product  of  the  fraction 
multiplied  by  itself  a  certain  number  of  times ;  and  a  fraction  is  mul- 
tiplied by  itself  by  multiplying  the  numerators  and  denominators  to- 
gether separately. 


EXAMPLES. 


1.  Find  the  — ^^  power  of  a".  Ans.  a  '  . 

2.  Find  the  m^^  power  of  a  ' .  Ans.  a~. 

3.  Find  the f' power  of  a".  Ans.  a~^ 


EXTRACTION     OF     ROOTS.  231 

l_  — m 

4.  Find  the  — m'"  power  of  a  ' .  ^4718.  a 

5.  Find  the  4*"  power  of  oaVi^  Ans.  81a^c'-6'«, 

6.  Find  the  3"  power  of  the  4<*  power  of  j-.       Ans.  12Sa''b-''. 

2a 

7.  Find  the  2'"  power  of  ^.  Ans.  (2)''a''t~^ 

8.  Find  the  —Shi*"  power  of  a^b^  Ans.  a-x^Z/-'^" 

9.  Find  the  — 5m^^  power  of  the  S"*  power  of  a^P. 

Ans.  a-^-^b-^" 


10.  Find  the  — 12«''  power  of  the  — G*"  power  of ^ 


+  a''b-''. 


11.  Find  the  — 12*"  power  of  the  — T""  power  of j-. 

Ans.    +a'*b-^. 

12.  Find  the  — 11"'  power  of  the  — 7""  power  of  — — . 

A71S.  —aPb-"". 

13.  Find  the  ?■">  power  of  the  ^n'"  power  of  a".  Ans.  a°°". 

14.  Find  the  — r*""  power  of  the  m^^  power  of  a~° 

15.  Find  the  r*"  power  of  the  — m^^  power  of  a~" 

16.  Find  the  — r*"  power  of  the  — m*"*  power  of  a° 

17.  Find  the  S""  power  of . 


18.  Find  the  5'"  power  of . 


19.  Find  the  C"  power  of  2a^b^. 

20.  Find  the  6"'  power  of  ~2a}b^ 

21.  Find  the  31  power  of  laH^. 


An. 

.  a'^^ 

An. 

.  a""'. 

An 

\  a'""^ 

Ans. 

ah* 

C5     • 

ns.  + 

C5 

Ans. 

Ua'b. 

Ans. 

Ua'b. 

Ans. 

8a6^. 

232  FORMATION    OF    THE    POWERS    AND 

1    1  1 

22.  Find  the  3''  power  of  —2a^h^.  A^is.  —Sab-. 

23.  Find  the  i""  power  of  Ma  "Lp.  Ans.  (Myall v. 

EXTRACTION    OF    ROOTS    OF    QUANTITIES    AFFECTED  WITH 
ENTIRE  OR  FRACTIONAL  EXPONENTS.  —  MONOMIALS. 

290.  To  extract  the  m'**  root  of  any  quantity,  a^,  is  to  find  a  second 
quantity  which,  when  raised  to  the  m'^  power,  will  produce  the  given 
quantity,  «■'.  The  extraction  of  a  root  is  then  just  the  reverse  of  raising 
to  a  power,  and,  of  course,  the  reverse  process  must  be  pursued  to  find 

p  p 

the  root.     The  m^^  root  of  a^  must  be  am,  because  the  m"*  power  of  am 

is  equal  to  a^.  To  raise  a  quantity,  which  has  a  coefficient,  to  a  power, 
we  first  raise  the  coefficient  to  the  indicated  power,  and  write  the  new 
coefficient  before  the  literal  factors  as  the  coefficient  of  the  power. 
Hence,  to  extract  the  root  of  a  quantity,  which  has  a  coefficient,  we 
must  first  extract  the  root  of  the  coefficient,  and  write  it  before  the  root 
of  the  literal  factors.  The  on'^  power  of  Ma^  is  (M)"'a''°',  therefore, 
the  m""  root  of  (M)"af""  must  be  Ma''.     The  ru*  power  of  the  fraction 

-  =  — .     Hence,  the  wi""  root  of  -, —  =  — .     That  is,  the  m""  root  of 

numerator  and  denominator  must  be  taken  separately. 


Extract  the  root  of  the  coefficient,  and  write  the  result  be/ore  the  lite- 
ral/actors affected  with  exponents  equal  to  the  quotient  arising  from 
dividing  their  primitive  exponents  hy  the  index  of  the  root.  If  the  co- 
efficient of  the  given  monomial  is  not  a  perfect  power  of  the  degree  of 
the  root  to  he  extracted,  the  operation  is  impossible.  If  the  exponents 
of  the  literal  factors  are  not  divisible  by  the  index  of  the  root,  the  lite- 
ral factors  win  appear  in  the  root  with  fractional  exponents.  The  root 
of  fractions  is  extracted  by  extracting  the  root  of  the  numerator  and 
denom  in  a  tor  separately. 

EXAMPLES. 

1.  Find  the  -"^  root  of  a  ^  Ans.  a". 

r 

m  Ans    (Z— 

2.  Find  the  r""  root  of  a^.  '     '^' 


EXTRACTION    OF    ROOTS.  233 


3.  Find  the  m'"  root  of  a^  Ans.  ar . 

1  " 

4.  Find  the '"  root  of  a'.  Ans.  a-°. 


5.  Find  the  — m'"  root  of  a  ■•  .  Ans.   a    ^  . 

6.  Find  the  —  j-'"  root  of  a  "^  .  Ans.  a~~^. 

m  I 

7.  Find  the  — m'"  root  of  a"  r  .  Ans.   a'r . 

1                          m  ^ 

8.  Find  the '"  root  of  a~  '  .  Ans.  a '  . 

m 

9.  Find  the  4'"  root  of  Sla'c'W.  Ans.  3aVt^ 

10.  Find  the  7'"  root  of  128a''i-''.  Ans.  2ah-\ 

11.  Find  the  q"^  root  of  (2)''a<'i-^.  Ans.  2ah-\ 

12.  Find  the  — Sm'"  root  of  a-«^t-'^.  Ans.  aW. 

13.  Find  the  r*"  root  of  a"".  Ans.  a"". 

14.  Find  the  4'"  root  of  l^Jn^.  Ans.  2a^^b^. 

rt"-/)3  a- 6^ 

15.  Find  the  5'"  root  of  —-.  Ans.    . 

c^  c 

5     5  11 

16.  Find  the  5*^  root  of  —  — ^.  Ans. . 

c"  c 

17.  Find  the  3''  root  of  8ah^.  Ayis.  2ah^. 

18.  Find  the  6'"  root  of  Mci'b.  Ans.  2a*6*. 

19.  Find  the  3*  root  of  — 8aZ>i  Ans.  — 2a*6*. 

t   21^  j  2 

20.  Find  the  t^^  root  of  (Mya^bp.  An-s.  Ma°ip. 


PROMISCUOUS   EXAMPLES. 

-i    -i    1  _3      _  3 

1,  Raise  2a  "b  ^c^  to  the  3*  power.  Ans.  Sa  -b    ic. 

_3    _£  _i  _j_  i 

2.  Extract  the  3"  root  of  8a  ^5    p.  Ans.  2a  'b    pc 

20* 


234  FOEMATION    OF    THE    POWERS    AND 

3.  Multiply  2a-P6-°'c-''  by  ZaPh'^c''.  Ans.  6 

4.  Divide  6  by  3a^h'^c\  Ans.  2a-Pi-"c-". 

5.  Divide  6  by  2a-P6-"'c-».  Aris.  Sa^h'^c^ 

6.  Raise  2a^b^c^  to  the  6'"  power.  Ans.  64a'&c^. 

7.  Extract  tlie  6'"  root  of  Q^a^bS.  Ans.  2a^b^c^. 

1       1  m— 1     n— 1 

8.  Multiply  (M)mar  by  (M)  ma-.  Ans.  Ma. 

m-l     E-1  Jl_     1 

9.  Divide  Ma  by  (M)"^a"^.  Ajis.  (M)'nan. 

1       1  in— 1     n— 1 

10.  Divide  Ma  by  (M)^an.  Ans.  (M)"^a~. 

11.  Multiply  8a;"V'  by  ^.  Ans.  xy. 


12.  Divide  a;y  by  8a   ^y  "*.  ^ns. 


4     5 
3,,4 


a;oy 


4     5 
3  ,,4 


13.  Divide  cry  by  ^^.  ^ns.  8a;  ^  ^. 

o 

14.  Raise to  the  third  power. 

Ans.  —Vlhx-^y-^z-^a~''b-^. 

15.  Raise  Va^V  to  the  r*  power.  Ans.  (Vya^'b^'. 

16.  Extract  the  r'"  root  of  (P)V^6^  ^ws.  Va%\ 

17.  Divide  (2)Vi=' by  (2ya'"i-".  ^ns.  2a' -"'J^'". 

18.  Multiply  2a''-"i2-m  itjy  (^2ya"i».  Ans.  Sa'b'. 

19.  Multiply  2a^'"&2-m  ^^^  (2)V"'J="".  ^?js.  (2)V"+'">J2('+'">. 

20.  Extract  the  s*"  root  of  (W)'^a-'b~'^ .  Ans.  (W)''a-'i""^. 

Li  10    20 

21.  Raise  3a^6'-  to  the  10*^- power.  Ans.  (py°a'b'. 

10   20  12 

22.  Extract  the  10"^  root  of  (3)"'a^6^  Ans.  Sa^b~'. 

23.  Multiply  (4)'a^^*  by  (ifah^.  Ans.  (4)V5. 


EXTRACTION    OF    ROOTS.  235 

24.  Divide  (4)V6  by  {^fah^.  Ans.  (4)'a^6*. 

25.  Divide  (4)Vi  by  (4)'a-6*.  Ans.   {Afah^. 

11  3     3 

26.  Raise  — a"&^  to  the  3''*  power.  Ans.  —  d^h^. 

i   I  4 

27.  Raise  — a'h^  to  the  4*"  power.  Ans.  +  a^i^. 

I      2     3 

28.  Raise  tx^y^z'^  to  the  c*"  power.  ^ns.  (2yxy^^. 

2     2    £ 

29.  Extract  the  c*"  root  of  (2)'x/^'.  ^ns.  Ix^y^z^. 

30.  Find  the  r'"  root  of  ^X.  Ans.  2a~^. 

31.  Raise  2a~'  to  the  r""  power.  .4hs.  (2)''  a"". 

32.  Raise  aVrUo  the  10'"  power.  Ans.  a%\ 

33.  Extract  the  10""  root  of  a'1/.  Ans.  a'b\ 

34.  Raise  a-^i«  to  the  2''  power.  ^tis.  ah''' 

35.  Extract  the  square  root  of  aZ*'- 1  vl?«s.  a-^i-®. 

36.  Raise  a-'^h-'^  to  the  6""  power.  Ans.  a-°"b-°'\ 

37.  Extract  the  6'"  root  of  a-o'^i-^'l  Ans.  a'^'b-'^. 

38.  Raise  a'"«6-«»  to  the  1000'"  power.  ^Ins.  a'b\ 

39.  Extract  the  lOOO'"  root  of  a'b^.  Ans.  a-°°'b-<^. 

40.  Multiply  ah^  by  a^^^i-^.  A^is.  a-'=i« 

41.  Divide  a-"6"^  by  a-6".  ^?is.  a-^b-^. 

42.  Divide  a-^5Z>-«  by  a-^S-^'.  ^ns.  a'i-*. 

43.  Multiply  (10)2a-2i-=' by  (lO/a- "J- ^  Ans.  (lO/a-^S-*. 

44.  Divide  (10)^a-^6'5  by  (\^fa=b\  Ans.  (\^fa-%-\ 

45.  Divide  (lOya-^i-^  by  (lO/a-26-3.  Ans.   (l^ya-^b-\ 

46.  Multiply  (10)-'a-'6-«by(10)'a--'6-^.  Ans.  (\Ofa'b-\ 


236  FORMATION    OP    THE    POWERS    AND 

47.  Raise  2aZ>V  to  the  yVh  power.  A71S.  (^ly'a-'hh". 

48.  Extract  the  yigth  root  of  {^y'a-'h-h^  Ans.  2ah^c\ 

49.  Raise  (2)':r?/V  to  the /^ih  power.  Ans.  {2y^x-^yh-\ 

50.  Raise  (2)5a'"5"  to  the  /gth  power.  Ans.  2a-^'"h-^\ 

It  will  be  seen  that  decimal  powers  are  smaller,  and  decimal  roots 
greater,  than  the  quantities  themselves. 

CALCULUS  OF  RADICALS. 

291.  Any  quantity  with  a  radical  sign  is  called  a  radical  quantity, 
or  simply  a  radical.  Thus,  \/a,  \^b,  Vc^,  are  radical  quantities,  or 
radicals. 

The  coefficient  of  a  radical  is  the  quantity  prefixed  to  the  radical 
sign,  and  it  indicates  the  number  of  times,  plus  one,  that  the  radical 
has  been  added  to  itself.  Thus,  2y/a  and  m^/a,  indicate  that  ^/ a  has 
been  added  to  itself  once  and  vi  —  1  times.  When  no  coefficient  is 
written,  unity  is  understood  to  be  the  coefficient.  Thus,  >/  a  =  ly/a. 
When  the  indicated  root  can  be  exactly  extracted,  the  radical  is  said  to 
be  commensurable  or  rational.  Thus,  ■v/4  and  ^8  are  rational  radicals. 
When  the  indicated  root  cannot  be  exactly  extracted,  the  radical  is  said 
to  be  incommensurable  or  irrational.  Thus,  ■s/2  and  ^  5  are  irrational 
or  incommensurable  radicals. 

A  root  has  been  defined  to  be  a  quantity,  which,  taken  as  a  factor 
a  certain  number  of  times,  will  produce  the  given  quantity.  An  even 
root  is  a  quantity,  which,  taken  as  a  factor  an  eve7i  number  of  times, 
will  produce  the  given  quantity.  But  no  quantity  taken  an  even 
number  of  times  will  produce  a  negative  result.  Hence  the  even  root 
of  a  negative  quantity  is  impossible.  The  indicated  even  roots  of 
negative  quantities  are  called  imaginary  quantities.  Thus,  y — 2, 
^ — 2,  v' — 2,  are  imaginary  quantities.  Roots  that  are  not  imaginary 
are  called  real  roots. 

The  term  rational  is  in  contradistinction  to  irrational. 

The  term  real  is  in  contradistinction  to  imaginary. 

A  quantity  may  be  real  and  not  rational ;  but  no  quantity  can  be 
rational  or  irrational  and  not  be  real.  Thus,  V2  is  real,  but  not 
rational;  but  v'4  and  \/2  are  both  real. 


EXTRACTION    OF    ROOTS.  237 

292.  It  has  been  shown  that  all  radicals  may  be  changed  into  paren- 
thetical expressions ;  the  numerator  of  the  exponent  of  the  parenthesis 
denoting  the  power  to  which  the  quantity  under  the  radical  sign  is 
raised,  and  the  denominator  of  the  exponent  of  the  parenthesis 
denoting  the  index  of  the  radical,  or  the  degree  of  the  root  to  be  ex- 
tracted.    Thus,   v/a*   may  be  changed   into   a-,    and    -(/a™  may  bo 

changed  into  a". 

293.  A  simple  radical  is  one,  which,  when  changed  into  an  equiva- 
lent parenthetical  expression,  has  unity  for  the  numerator  of  the 
exponent  of  the  parenthesis.  Thus  ^a,  i/a,  &c.,  are  simple  radicals. 
A  complex  radical  is  one  which  has  the  quantity  under  the  sign  raised 
to  some  power  different  from  unity.     Thus,  \/a',  ^a",  are  complex 

3  ^ 

radicals,  the  equivalent  parenthetical  expressions  («)"-,  and  (a)",  having 

numerators  different  from  unity. 

29-4.  Radicals  are  said  to  be  similar  when  they  have  the  same  index 

and  the  same   quantity  under  the    sign.     Thus,  2yja  and  3's/a  are 

similar  radicals.     The   ^  a  and   ^a  are  not  similar,  because,  though 

the  quantities  under  the  radical  signs  are  the  same,  the  indices  of  the 

radicals  are  different.     The   s/ a  and   \/6  are  dis.similar,  because  the 

quantities  under  the  signs  are  different. 

295.  Similar  powers  of  the  same  quantity  can  be  added  by  adding 
their  coefficients.  Thus,  2a^  -f  3a^  =  5a^ ;  because,  since  the  literal 
factors  are  the  same,  they  may  be  represented  by  the  same  letter,  x. 
Then,  2a*  -f  3a^  =  2x  +  3x  =  Sx,  and  replacing  x  by  its  value  a^,  we 
have  the  sum  of  the  two  quantities  equal  to  ba^.  For  a  like  reason, 
similar  powers  of  the  same  quantity  may  be  subtracted  from  each  other, 
by  taking  the  difference  between  their  coefficients  and  uniting  this 
difference  as  the  coefficient  of  the  common  quantity.  Thus,  5a'  — 
2a»  =  3a^ 

296.  In  the  same  way,  similar  radicals  may  be  added  or  subtracted, 
2s/ a-{-  •y/a  =  3v/a.  For,  make  V a  =  a; ;  then  2v/a-|-  v/«  =  2x  + 
X  =  Zx  =  Z^ a.     So,  also,  2s/a —  y/a  =  y/a. 

297.  The  calculus  of  radicals  shows  how  radicals  may  be  operated 
upon  algebraically — added,  subtracted,  multiplied,  &c.,  &c. 

These  algebraic  operations  must  be  in  accordance  with  certain  prin- 
ciples, which  both  modify  and  facilitate  them.  It  would  require  an 
extended  treatise  to  embrace  all  the  principles  of  the  calculus  of 
radicals.     A  few  of  the  most  important  only  are  given  in  this  work. 


238  FORMATION    OF    THE    TOWERS    AND 

First  Principle. 

298.  Any  parenthetical  expression,  composed  of  several  factors,  may 
be  decomposed  into  as  many  new  parenthetical  expressions  as  there  are 

p  p        p       p 

factors.     Thus,  (aZ>c)T  =  (a)q  (5)q  (c)q  . 

For  by  the  rules  for  raising  a  monomial  affected  with  any  exponent, 

fractional  or  entire,  to  a  power,  each  factor  has  to  be  separately  raised 

P  P  £  P 

to  the  indicated  power.     Hence,  (cihc)i  —  (a)q  (b)i  (c)q  .     And,   in 

p  P  L         H 

like  manner,  (a'"i"c'')q  =  (a'")q'  (6")i  (c')''  • 

299.  This  principle  has  an  important  application  in  the  case  of  simple 

I I        I        1         I 

radicals.     For  (abed)'"    =  y  abed,  is  also  equal  to  (a)"  (&)n  (cf  {dy 

or  to  ^a  .  y  6  .  ^c  ^ d.  Hence,  ^ abed  =  </a  ^b  ^c  ^ d.  That 
is,  the  n""  root  of  the  product  of  any  number  of  factors  is  equal  to  the 
product  of  the  n"*  roots  of  these  factors. 

300.  This  property  of  radicals  is  used  for  two  distinct  purposes. 
1st.  To  simplify  radicals.     Let  it  be  required  to  simplify  or  reduce 

V  75a^c.  We  see  that  the  quantity  under  the  sign  Iba^c,  can  be  de- 
composed into  two  factors,  25a^  and  3c,  and  that  the  first  is  a  perfect 
power  of  the  degree  of  the  root  to  be  extracted.  Then,  V  l^a^c  = 
v'25a^  X  3c,  which  is  also,  by  the  property  just  demonstrated,  :=: 
v/25^2  v/3^  =  5a  V3Z  In  like  manner,  ^l^^c  =  V^^  X  ^7= 
2rt  ^c.     Then,  to  simplify  or  reduce  a  radical,  we  have  the  following 

RULE. 

Decompose  the  quantity  under  the  sign  into  two  factors,  one  of  which 
shall  be  a  perfect  poioer  of  the  degree  of  the  root  to  be  extracted.  Ex- 
tract the  root  of  the  perfect  power,  and  write  this  root  before  the  incom- 
mensurable factor. 

301.  2d.  The  property,  that  the  li""  root  of  the  product  of  any  number 
of  factors  is  equal  to  the  product  of  the  n""  roots  of  those  factors,  is  also 
used  to  make  radicals  similar  which  appear  dissimilar.  Then,  after 
they  have  been  made  similar,  they  may  be  operated  upon  algebraically, 
added,  subtracted,  &c.,  like  simple  algebraic  expressions.  

Thus,  -J  ^a^c  and  m-J  c  appear  to  be  dissimilar.  But  y/ 4:a^c  =^ 
\/4a^ .  c  =  \/4«^  -y/c  =  2iasf  c.  Hence,  if  it  be  required  to  add  \/\d^c 
and  m  v"  f ?  we  may  represent  %/  c  by  .r,  and  the  sum  of  the  two  quantities 


EXTRACTION    OF    ROOTS.  239 

will  be  2ax  +  mx  or  (2a  +  m)  x.  Now  replace  x  by  its  value  -J  c,  and  we 
have  ^/ia^c  -{■  m-J  c  =^  (2a  +  m)-sfc.  In  like  manner,  26^8aV 
and  — 2ac^c^V^,  appear  to  be  dissimilar.  But,  26y8aV  = 
2b^Sa^c'  x~?  =  26^8^  Xj/7  =  ^achV^,  and  —  lacVTV'  = 
—  2ac^  6^(c^)  =  —  2ac  .e^  6«  V  c^  =  —  2ao6  ^7.  Hence,  26  ^  8aV  — 
2ac^<?b^  =  2ac6v/c^.  To  operate  upon  dissimilar  radicals,  we  bave 
then  the  following 

RULE. 

First  make  the  radicals  similar,  then  represent  the  common  radical 
hy  X,  or  conceive  it  to  he  so  represented,  and  operate  upon  the  new  radi- 
cal expressions  according  to  the  rules  for  simple  algebraic  expressions. 

The  following  examples  will  afford  illustrations  of  the  two  applica- 
tions of  the  principle,  that  ^abc,  &c.  =  ^a .  ^b  ^ c,  &c.  For  the 
sake  of  simplicity,  we  will  assume  the  own  signs  of  all  the  radicals  to 
be  positive. 


1.  Simplify  the  expression  v/lGaV6*.  Ans.  AacV^y/cb. 


2.  Simplify  the  expression  ^'llxS/z'.  Ans.  Sxy^z^s/z. 


3.  Simplify  the  expression  V50a'6V.  Ans.  bab\'.X/2ahc. 


4.  Simplify  the  expression  y24a*6V.  Ans.  2a6Vy3a" 


5    Add  the  radicals  V16a='6"'c2  and  +  2a¥c^a. 

Ans.  Qah^Cy/a. 


6.  Add  the  radicals  v/ SOa^^i^"  and  +  5a'"Z;^V26-^°. 

Ans.  10a""i2ny2^ 


7.  Add  the  radicals  4  V  a-^fc-^c-P  and  +  oa'^'V'c^^ ^ a'H-^-'''^ . 


Ans.  1  s/ar'b-'c-^. 

\.  Add  the  radicals  4:a'b'^  ^/'?'  and  8c  vV^. 

Ans.  Vla'b^sj'c'. 

).  Reduce  -J 21  +  s/i2  —  -Jlh  to  one  sum. 

Ans.  Ov/37orO. 


240  FORMATION    OF    THE    POWERS    AND 

10.  Reduce  750"+  n/8+  v/32"+  \^18  +  n/162  to  one  sum._ 

Ans.  23  v/ 2. 

11.  Reduce  2^^  +  4^7"+  2ay7to  one  sum. 

Ans.  (2  4- 4c  +2ac2)^'c^ 

12.  Reduce  ay/ 20  +•  Jv/5  —  Cs/bU^  to  one  sum. 

yIhs.  (2a  +  &  — c&)v/5'. 

13.  Reduce  ab\/ 72  +  c\/8  +  m«\/18<-  to  one  sum. 

Ans.  (Gah  +  2c-\-Smnc)y/^. 

14.  Reduce  5  yS'and  ^SaV  to  one  sum.         ^l«s.  (5  +  ac2)4/8. 


15.  Reduce  12  V"'^"^  and  m^d'.  Ans.  (I2ab^  +  md)^d. 


16.  Reduce  a  ^  6='+V+^?i  and  m  .y  n'+'6V.  

J.ns.   (ahc -{■  mn')  -^b^c^n. 

17.  Reduce  ViS+SaN/T^  +  e^vMS  — 24JVT2. 

Ans.  (2  +5rt  — 12Z>)x/12. 

18.  Reduce  40^^"  +  SacVc"^  —  9ac\/c  to  one  sum.        Ans.  0, 

19.  Reduce  12iV«"'"'"c'°'  +  5m  V«  to  one  sum. 

Ans.  (12&ac2  +  5m)  Va^ 

20.  Reduce  3\/aV6^  and  — 2abcy/abc  to  one  sum. 

^?is.  abcy/abc. 

The  whole  difficulty  in  solving  these  examples  consists  in  finding  the 
incommensurable  factor  common  to  all  the  radicals. 


Second  Principle. 

302.  The  Ji""  root  of  the  quotient  of  two  quantities  is  equal  to  the 

quotient  of  their  w*  roots.     That  is,  \  /  —  =  — — 

V    6         «/6- 

For,  to  raise  a  fraction  to  any  power,  we  raise  the  numerator  and 
denominator,  separately,  to  the  required  power.  Hence,  to  extract  any 
root,  as  the  n'",  the  root  of  each  term  of  the  fraction  must  be  extracted 
separately. 

This  principle  is  used  in  extracting  the  roots  of  fractions. 


EXTRACTION    OF    ROOTS.  241 

Third  Principle. 

303.  The  mn^^  root  of  any  quantity  is  e(|ual  to  the  m^^  root  of  thy 

n^  root  of  that  quantity.     That  is,  '":y~a  =  k/  ^ya. 

A  quantity,  a,  may  be  raised  to  the  sixth  power  by  first  squaring  a, 
and  then  cubing  the  result.  Hence,  the  sixth  root  of  a®  might  be 
extracted  by  first  extracting  the  square  root,  and  then  extracting  the 
cube  root  of  the  result.  So,  a  may  be  raised  to  the  wiJi""  power  by 
first  raising  it  to  the  ?<'"  power,  and  then  raising  the  result  to  the  m"" 
power.  Hence,  the  m?*"'  root  can  plainly  be  extracted  by  taking  first 
the  n"",  and  then  the  m""  root,  in  succession. 

304.  This  principle  is  of  great  importance,  and  extensive  applica- 
tion. It  is  used  to  extract  high  roots  in  succession,  whenever  their 
indices  can  be  decomposed  into  factors. 


Thus,  V256(W^=:\/y"2o 


m^iW  =  ^  l()a^6«  =  \a^V 


All  even  roots,  and  roots  that  are  multiples  of  3,  can  be  extracted  in 
succession.  The  sixth  root  is  equal  to  the  cube  root  of  the  square  root. 
The  eighth  root  is  equal  to  the  fourth  root  of  the  square  root,  or  it  can 
be  found  by  extracting  the  sfiuare  root  three  times.  But  the  fifth  root, 
seventh  root,  eleventh  root,  &c.,  cannot  be  extracted  successively. 

When  the  principle  can  be  applied,  we  begin  by  extracting  the  low- 
est root  first.  The  index  of  the  radical  must  be  decomposed  into  its 
prime  factors,  and  the  root  corresponding  to  the  lowest  factor  ought  to 
be  first  extracted.     It  frequently  happens  that  the  factors  are  equal. 


EXAMPLES. 

1.  Extract  the  eighth  root  of  256a'i^ 


2.  Extract  the  sixth  root  of  Tl^a}%^\ 


3.  Required  V  6561  a^^6'2  j^^g    g^ejs^ 

4.  Required  ^'25<o^F'.  Ans.  4:ah^. 
21                                           Q 


242  roRMATioN   or  the  towers  and 


1  1 


5.  Required  V4096a6.  Ans.  4a^i®. 

8/  4~T  1    1 

6.  Required  \765536a^6^  Ans.  4a%^. 

7.  Required  -^/MOl^W^  Aiis.  Sa^L\ 

8.  Required  "^  a'^h'K  Ans.  ah\ 

9.  Required  ^-^IZly^iHj^.  Ans.  2,a^l\ 
In  this  example,  extract  the  s""  root  first. 

Fourth  Principle. 

305.  The  coefficient  of  a  radical  may  be  passed  under  the  radical 
sign  by  raising  it  to  a  power  indicated  by  the  index  of  the  radical. 
That  is,  PVa  =  7P"a.  For,  P^yP^;  hence,  P V«'=  7P"  7'"^; 
but,  by  the  first  principle,  V~^  7^=  VP"a-  Therefore,  P7a  = 
VP'^ 

306.  The  fourth  principle  is,  obviously,  just  the  converse  of  the 
first  principle.  The  latter  enables  us  to  pass  a  factor  outside  of  the 
radical ;  the  former,  to  place  a  factor  under  the  radical.  The  fourth 
principle  is  frequently  used  in  the  differential  calculus.  It  has  two 
applications  in  Algebra:  1st.  The  approximate  value  of  incommen- 
surable roots  can  sometimes  be  found  more  exactly  by  means  of  this 
principle.  Thus,  |n/T=  n/(|)'^.  7  =  V^^s  =  | :  true  to  within  less 
than  \ .  A  nearer  approximation  to  the  value  of  the  expression  has 
been  found  in  this  instance  by  passing  the  coefficient  under  the  radical. 
But,  for  such  expressions  as  |  ^/  2,  f  ^/  5,  nothing  is  gained  by  passing 
the  coefficient  under  the  radical. 

2d.  The  fourth  principle  may  be  used  instead  of  the  first,  or  in  con- 
nection with  the  first,  in  making  radicals  similar  which  appear  dissimi- 
lar. Thus,  9av/6,  and  v'^lo^  are  dissimilar,  until  either  9a  is 
passed  under  the  radical,  or  81a^  is  passed  without.  Sometimes  it  is 
difficult  to  employ  the  first  principle  to  make  radicals  similar,  because 
it  is  not  easy  to  perceive  the  incommensurable  factor  common  to  all  the 
radicals.  But  there  is  no  difficulty  in  employing  the  fourth  principle. 
Pass  all  the  coefficients  under  their  respective  radical  signs,  and  then 


EXTRACTION    OF    ROOTS.  243 

decompose  the  numerical  and  literal  factors  into  their  prime  factors. 
If  the  radicals  can  be  made  similar,  common  factors  will  become  appa- 
rent after  the  decomposition  into  prime  factors.  Let  it  be  required  to 
ascertain  whether  2a  ^ab^,  and  h^a^b^  are  similar.  The  equivalent 
expressions  are  ^Ha'^b^,  and  ^  a^b^.  The  two  radicals  have,  then,  a 
common  factor,  a^b^,  under  the  sign.  This  common  factor  can  be  de- 
composed into  two  others,  one  of  which  is  a  perfect  power  of  the  degree 
of  the  root  to  be  extracted,  and  the  other  incommensurable.  The  in- 
commensurable factor  may  be  made  a  new  radical,  and  the  given  ex- 
pressions being  decomposed  into  factors,  one  of  which  is  this  incom- 
mensurable factor,  will  be  similar.  We  have,  in  accordance  with  this 
rule,    2ay^b'=  ^J^  -^  ^"8^^  ^IdJ^  =  2ab^7d/,    and    bV^^ 

=  V^^=  ^'^^ab'  =  ab^ab\  

Take,  as  a  second  example,  7  V  x^,  and  x  ^24:01x1/  ;  by  passing  7 
and  X  under  the  radical  signs,  we  have  ^2A0\x^if,  and  v/2401a;*^. 
Hence,  V2401a;^  is  a  common  factor,  and  the  given  expressions  reduce 
to  Ixi/ xif  and  lx\/ xy,  and  are  not  similar,  since  they  have  no  com- 
mon incommensurable  factor.  We  have,  then,  a  simple  process  for 
ascertaining  whether  radical  expressions  can  be  made  similar. 


RULE. 

Pass  the  coefficients  of  the  different  radicals  tinder  their  respective 
signs,  and  then  examine  whether  there  is  a  common  incommensurable 
factor.  Jf  so,  the  radicals  vxay  be  made  similar  by  taJcing  the  incom- 
mensurable factor  as  a  new  radical.  Write  before  it,  for  new  coeffi- 
cients, the  roots  of  the  commensurable  factors  of  the  respective  expres- 
sions, and  we  loill  have  a  series  of  new  radical  expressions,  all  similar. 
If  there  is  no  common  incommensurable  factor,  the  radical  expressions 
are  not  similar. 

EXAMPLES. 

1.  Add  together  2a  ^Jc^  and  ^c¥jb?. 


Then,  2aybc^=  V4:a^_bf=  sf\a\^>Jbc  and  ^cb^s/b&=  >/96V  = 
s/^V&  -Jbc.  Hence,  sfbc  is  a  common  incommensurable  factor,  and 
the  sum  is  {2ac  -f-  35c)  s/  be. 

2.  Add  Ax'^y^s/a^<?  and  aczs/a¥c  together. 

Ans.  i^do^y"^  -f  b^z)  ac-f  ac. 


244  FORMATION    OF    THE    POWERS    AND 

3.  Add  toi^ether  J^SiO^^  and  Gas/ 240a. 


h  ^/  540o'>  =  y/  3 . 5  ■2^3V6^  and  6a  V  240a  =  V  36 . 3 . 5 .  4V. 
Hence,  y/o.S.a,  or  \/15a,  is  a  common  incommensurable  factor,  and 
the  sum  of  the  two  expressions  is  (Qah  +  24a)  \/ 15a. 

4.  Add  together  V90  and  V'MO. 


y/90=V3^5.2  and  v/810  =  n/3\2.5.     Hence,  v/2.5  is  the 
common  incommensurable  factor,  and  the  sum  is  (3  +  9)v/  10. 

5.  Add  together  as/2AQ  and  Z>n/2400.     ^hs.  (6a  +  206)  VG". 

6.  From  a  v/300  subtract  Z>V"243.  Ans.  (10a  — 9Z>)V37 


7.  From  a  V  33750  subtract  6-^/10000. 

Ans.  (Sa  —  2h)  ^l250" 

8.  Add  the  three  expressions,  hay/ ax,  mx^\/Qa'^,  na"  s/ x. 

Ans.   (ha  +  3ma  +  n^s/ax. 

9.  Add  the  three  expressions,  ha^~ax,  mx'^^27a*  +  na^  ^x. 

A?is.   {ha  +  3ma  +  n)^ax. 

'ax  +  mx*  ySla^  +  na'^ ^Ic. 
Ans.  (ha  +  Sma  -f-  n)^ax. 


10.  Add  the  three  expressions,  haijax  -{■  mx'^  y  81a^  +  na'^  ij x. 


Fifth  Principle. 

307.  Any  factor  may  be  passed  without  a  parenthetical  expression, 
by  multiplying  its  exponent  by  the  exponent  of  the  parenthesis. 

p  mp  p 

Thus,  (a'"hc)  q  =  a  1  (6c)<i . 

L         1LP  Jl  P      .  n 

For,  (a"'hc) i  =  a'ih'ic'i,  since  each  term  has  to  be  raised  to  the  — 

9 

nip  p    p 

power.     And,  by  separating  into  factors,  we  have  at  x  tqci,  which  is 

mp  p 

also  equal  to  a  •)  (hc)^ . 

If  there  is  more  than  one  term  within  the  parenthesis,  any  term  or 
a  factor  of  any  term  may  be  passed  out,  by  dividing  all  the  terms  by 
the  expression  to  be  passed  out,  and  then  writing  that  expression  out- 
side of  the  parenthesis,  with  its  primitive  exponent  multiplied  by  the 


EXTRACTION    OF    ROOTS.  245 

exponent  of  the  parenthesis.     Let  it  be  required  to  pass  a"  without  the 

parenthesis  (a^S  +  c)i.     This  may  be  written  \a'°lb  +  —  I  p=  (a"'J)i 

^      p  g  p 

—  ai  (fZ)^.     In  which  d  =  h  -\ — ~  or  h  -{-  a-'^c     Hence,  (a"'Z>  +  c)5 

Dip  p 

It  matters  not  how  many  terms  there  may  be  within  the  parenthesis ; 
they  may  be  all  represented  by  a  single  letter,  after  they  have  been 
divided  by  the  factor  to  be  passed  without.  The  foregoing  demonstra- 
tion will  then  be  applicable. 

EXAMPLES. 

1.  Pass  X  without  the  parenthesis  (xy^z^)~~^. 

Ans.  x~^(i/^z^~ 

2.  Pass  x~^  without  the  parenthesis  (x~^y^z^y. 


Ans.  X   "  (i/^z^' 


Pass  X      without  the  parenthesis  (:c     ah)- 


Ans.  x(ahy 

4.  Pass  4  without  the  parenthesis  (4a2i')-.  Ans.  2(a2?'^)- 

-  £  5-  1 

5.  Pass  x'' without  the  parenthesis  (a:''?/'"^)P.  Ans.  x(y'";^y, 

6.  Pass  8  without  the  parenthesis  (Sahy.  Ans.  2  (ahy. 


7.  Pass  IG  without  the  parenthesis  (IGahy .  Ans.  2(ah) 


8. 

Pass  4  without  the 

.    (4«  +  ?0^-                / 
parenthesis  g .      Ans.  \a 

.r 

9. 

Pass  5  without  the 

parenthesis  (a 

+  5&)». 

0. 

Pass  2  without  the 

Ans. 

parenthesis  (2a  +  4iy. 

Ans. 

125(1 
16  (a 

+  2hy 

21* 


24G  FORMATION    OF    THE    POWERS    AND 

(4  (mx  +  nx"^)  +  (m  +  2nxf  )* 


11.  Pass  4  without  tlie  parenthesis 
Ans 


3 

(mx  +  ^ix"^  +  4  (wi  +  '2-nxy  )- 


12,  Pass  the  term  4a  without  the  parenthesis  (4a  +h  +  cy. 

Aiif;.   la   \i-+ ^ /   • 

13  Pass  the  term  8a'  without  the  parenthesis  (Sa"  +  i  +  0  • 


Ans.  2a\  o 


14.  Pass  the  term  bo?  without  the  parenthesis  (pci?  -{■  h  -\-  c)" . 

Any  number  of  terms  may  be  passed  without,  by  representing  them 
by  a  single  letter,  and  then  replacing  that  letter  by  its  value  after  the 
transfer  has  been  made. 

15.  Pass  (a  +  i)  without  the  parenthesis  (a  +  i  +  c)".  Let  a  -f  h 
=  X.  Then,  (a  +  h -\-  cf  =  (x  ■\-  c)*  =  x^(l  +  x-'c)^  =  (a+  i)" 
I  1  +  (a  +  i)-'c  1^ 

16.  Pass  (a  +  h  -\-  c)  without  the  parenthesis  (a  -{-  h  -\-  c  -}-  dy. 

.h 


Ans.  (a  +  i  +  «)-  I  1  +  (a  +  &  +  cy\l  \ ' 

17.  Pass  (a  +  i  4-  c)  without  the  parenthesis  (a  -\- h  +  cf. 

Ans.  (a  +  6  +  cy  (Tf  or  (a  +  &  +  cj. 

18.  Pass  a~°  without  the  parenthesis  (a""  +  ly. 

Ans.  a~p(l  +  a^hyp . 

The  last  example  shows  that  any  factor  affected  with  a  negative  ex- 
ponent may  be  made  to  appear  with  a  positive  exponent  in  the  other 
terms  of  the  parenthesis.  This  transformation  is  used  in  the  differential 
calculus  when  it  is  desired  to  change  a  negative  into  a  positive  exponent. 

It  is  evident  that  the  fifth  principle  is  only  the  more  extended  appli- 
cation of  the  first  principle. 


EXTRACTION    OF    ROOTS.  247 

Sixth  Principle. 

308.  Any  factor  may  be  passed  within  a  parenthesis,  by  multiplying 
its  exponent  by  the  reciprocal  of  the  parenthesis. 

Because,  to  pass  it  out  again,  we  must  multiply  its  new  exponent  by 
the  exponent  of  the  parenthesis,  and  when  its  new  exponent  has  been 
formed  as  directed,  the  factor,  after  it  has  been  passed  out  again,  will 

p  ^     p. 

be  afifected  with   its  primitive    exponent.      Thus,   a'"(i)'»  =(ai'Z>)'i. 

Because,  when  av    is  passed  out  again,  the  expression  will  become 

mq     p  2  p^ 

OP     q(i)<i  =  a'"(6)i. 

EXAMPLES. 


A71S.  I  {2ya  +  {2fb  I". 


1.  Pass  the  coefficient  2  within  the  parenthesis  (a  -f  U) 

2.  Pass  the  coefficient  2  within  the  parenthesis  (a  +  i)  . 

Ans.  (8a  +  U). 

—  — 

3.  Pass  a"  within  the  parenthesis  (a"  -f-  i)i . 

^q ,  q  »q      p_ 

Ans.  (a  p    p   +  api)q . 

4.  Pass  a"  within  the  parenthesis  (a    p  -|-  6)i . 

mq       p_ 

.4ns.  (1  +  a  p^»)i. 

5.  Pass  a""  within  the  parenthesis  (a""  +  H)     . 

Ans.  (a-^  +  Z*a-="")~*. 

6.  Clear  8(ia  +  iZ>)"-  of  its  coefficient.  Ans.  (^  +  A-. 

7.  Clear  64("--^  +  -^Y  of  its  coefficient. 

\4U9ba        lb/ 

8.  Clear  27(2a  +  |6)-  of  its  coefficient.  Ans.  (18a  +  b)^. 


248  FORMATION    OF    THE    POWERS    AND 

9.  Clear  32 (2a  +  Ib)^  of  its  coefficient.  Ans.  (8rt  +  b)^. 

Ans.  (25a  +  1256)" 


--a  +  -b\^  of  its  co- 
efficient. 


11.   Clear  tlie  parenthetical  expression  a~(l  +  a~')^  of  its  coefficient. 

Ans.  (a-j-  1)^ 

It  is  evident  that  the  sixth  principle  is  only  the  more  extended  ap- 
plication of  the  fourth  principle. 


Seventh  Principle. 

309.  The  denominator  of  the  exponent  of  a  parenthetical  expression  may 
be  multiplied  by  any  quantity,  provided  vre  raise  the  quantity  within 

p 
the  parenthesis  to  a  power  denoted  by  the  multiplier.     Thus,  (0)1  = 

Z  pp  JL         '^  -^  JLJL 

(a'")""'.    For  (a)T  =  an,  and  (a"')""!  =  W'l  =  a  ■) .   Hence,  (a)  i  =  (a'")°"i. 

If  there  is  more  than  one  term  within  the  parenthesis,  their  algebraic 
sum  may  be  represented  by  a  single  letter,  and  the  foregoing  demon- 
stration is,  therefore,  applicable  to  all  kinds  of  parenthetical  expres- 
sions. 

810.  The  seventh  principle  has  two  applications.  1st.  It  is  used  to 
cause  complex  radicals  (or  parenthetical  expressions  with  the  nume- 
rators of  their  exponents  different  from  unity),  to  be  alFected  with  ex- 
ponents having  a  common  denominator. 

Thus,  (a  +  by^  and  (a  -f  i)^,  can  be  changed  in  accordance  with  the 

3  4 

principle  into  the  equivalent  expressions  (  (<x  -f-  bf)*^  and  ((a-f  6)'^)^. 

We  see  that  a  second  parenthesis  has  been  written  within  each  pa- 
renthesis ;  and  we  see,  also,  that  the  exponent  of  the  first  new  paren- 
thesis is  the  quotient  arising  from  the  division  of  6,  the  least  common 
multiple  of  the  denominators  of  the  exponents  of  the  given  parentheses, 
by  the  denominator  of  the  exponent  of  the  first  given  parenthesis. 
The  exponent  of  the  second  new  parenthesis  is  the  quotient  arising 
from  dividing  the  same  least  common  multiple,  6,  by  the  denominator 
of  the  exponent  of  the  second  given  parenthesis.     In   like  manner, 


EXTRACTION     OF     ROOTS.  249 

2  3  .  -3_ 

(a  +  J)3  and    (a  +  5)*    may  be   changed  into    ((a  +  by)^'    and 

Hence,  to  cause  parenthetical  expressions  to  be  affected  with  expo- 
nents having  a  common  denominator,  we  have  the  following 


RULE. 

Take  the  least  common  multiple  of  all  the  denominators  of  the  ex- 
ponents of  the  parentheses,  and  divide  that  multiple  hy  the  denominator 
of  the  ej-ponent  of  every  parenthesis.  The  several  quotients  will  indi- 
cate the  power  to  which  the  quantity  within  their  respective  parentheses 
must  he  raised. 


1.  Change  [a  +  £)">  and  (a  H-  J)""",  into  equivalent  parenthetical  ex- 
pressions, the  denominators  of  whose  exponents  shall  be  the  same. 

P_  _£_ 

Ans.  (  (a  +  ^)°  )  '"''  and  (a  +  t)""". 
JL  -P. 

2.  Change  (a  -f  t)™*  and  (a  -f-  i)""-,  into  equivalent  expressions. 

Ans.  (  (rt  -f  hy  )  ■»"■'  and  (  (a  +  Z^)"  )  ■»^°. 

311.  2d.  But  the  most  important  application  of  the  seventh  prin- 
ciple, is  in  reducing  simple  radicals  to  a  common  index. 

*  ?.  —  —  J  -.  _ 

Thus,  since  a"  =  o*",  or  \/ a  =  ^a^,  and  a    =  a",  or  %/ a  =  -J/o', 

it  is  plain  that  \/a  and  ■i^a,  can  be  reduced  to  a  common  index.     So, 

^a  and  ^a,  may  be  changed  into  the  equivalent  radicals  '"^^a"'  and 

'"^a".     In  each  instance,  the  power  to  which  the  quantity  under  the 

radical  is  raised  is  indicated  by  the  quotient  arising  from  dividing  the 

least  common  multiple  of  the  indices  of  all  the  radicals  by  the  index 

of  the  radical  under  consideration. 


RULE. 

Form  the  least  common  mxdtiple  of  the  indices  of  all  the  radicals. 
Raise  the  quantity  under  each  radical  to  a  power  indicated  hy  the  quo- 
tient arising  from  dividing  the  least  common  multiple  hy  the  index 
of  the  radical  under  consideration. 


250  FORMATION    OF    THE    POWERS    AND 

Reduce  v'  a,  i/  «j  and  ^/  a,  to  same  index.  The  least  common  mul- 
tiple is  12,  and  the  three  quotiente^  ^'i^  ^>  V^  ^  ^>  ^°^  V'  =  ". 
Then  the  equivalent  radicals  are  -7  a*,  J^  a^,  and  '-^  al 

EXAMPLES. 

1.  Ecduce   y"a^  Va^  y«^  and  ^~a^io  same  index. 

^«s.  'V'^,  7"^,  7"^,  and  ^"^. 

2.  Eeduce  Va,  v/a,  ■C/c,  '^o,  and  v'a,  to  same  index. 

Ans.  ^-i/"^5^  ^"^2,  ^^^,  ^"^,  and  ^o^.' 

3:  Reduce  ^  a,  'X/a,  '-7 a,  V  a,  and  ^ a,  to  same  index. 

^ns.  ""iP/7i"",  '""^~a",  ""^a"',  ""Vo^,  and  "Va^. 

4.  Reduce  </a,  V«j  \^a,   ya,  and  Va,  to  same  index. 

12s' 


Ans. 


■^a'%  '^a'^  ■^.n  >^a»',  and  '^a'.' 


5.  Reduce  ^o,  ^  a,  ^^  a,  ^  a,  '-^a,  Va,  and  Sya,  to  same  index. 

Ans.    7^,  V~^,  7"<  7^  VTtV7^and  ^"^ 

6.  Reduce  '^/a,  "yo;  ^ct,  y/  a,  ^a,  and  '"-7  a,  to  same  index. 

Ans.  "^"^""^  -"7^^  "71:?°,  "70°^',  "^o^  and  '"7"(?^ 

7.  Reduce   -;/«,  ^"^  ^^  4/o7V«7  v' ^7  V^T  and  7«7  to  same 
index. 

^HS.     2520/-^^      2520,"^,      ^520,^     2520/-^4^     2520/-^0^     ^^^O/"^      2520/^ 

and  2520/^, 

It  will  be  seen  that  simple  radicals  are  changed  into  complex  by  the 
operation  of  reduction  to  a  common  index. 


Eighth  Principle. 

312.  Any  factor  of  the  index  of  a  complex  radical  may  be  sup- 
pressed, provided,  (he  same  factor  is  suppressed  in  the  exponent  of 
the  power  to  which  the  quantity  under  the  sign  is  raised.  That  is, 
"7  a"  =  V  a. 

For,  '"7  a'"  =  (a)"^  =  («)"  =  \^  a- 


EXTRACTION     OF     ROOTS.  251 

This  principle  is  just  the  converse  of  the  last,  and  reverses  the  results 
of  the  last.  It  is  used  to  simplify  complex  radicals,  and  frequently 
reduces  them  to  simple  radicals. 


EXAMPLES. 


1.  Simplify  Via  +  hf,  V (a  +  hj,  '^ (a  +  hf  and  l^(a  +  h)\ 
Ans.   >J  a  +  6,  V  a  +  6,   y  a  +  h,  and  ^  a  +  h. 


2.  Simplify  fi/(a  +  h)\  'Via  +  i)'°,  V{ci  -f  hf,  and  8/(a  +  i/. 
^«s.    v/(a  +  hf,  (a  +  ^')',  Va  +  i,  and  X/(a  +  6)'». 


3.  Simplify  v/(a+6)«,  V(a+t)',  !e/(a+&)'S  V(a+^')^  V(«+i)'*- 

^ns.   (a  +  />)^  ^(a  +  Z/),  N/(a  +  6/,  («  +  i),  (a  +  i)^ 

4.  Reduce  X/~^,  V<  ^ <  and  V'^. 

Ans.    V a,  i/c,  y/a,  and  y/a. 

5.  Reduce  i^y"^,  7~^,  7^,  V"^  and  7^ 

Ans.    i/'a,  i^i,  V«,  7"^,  and 7 oT 

6.  Reduce  ""Vo^,  "'"</aS  ""Vo^  ""V^^,  "V^i^- 

^7JS.   ^«,  Vfl,  7«,  yo,  and  -C^a. 

7.  Reduce  ^^^"^5,  ^^1?,  ^^7?^,  ^7^,  «y;?;  and  «yZ 

j47is.    v'a,  ^a,  7  a,  7«,  7«,  and  7  a. 

8.  Reduce  '"7«"S  '"'</~a",  ""7  a",  ""70^%  '"7'o^,  and  '"X/'a\ 

Ans.    7a,  "7a,  7o,  7a,  v'a,  and  '"7a. 

9.  Reduce  ^''^°/ a'""",  =^^7"^,  ^320/^^  2520/-^^  2520/"^  2520,  mT  mzo^"^^ 
and  237"^ 

Ans.    s/a,  7a,    7a,    7a,    7*0",  ^o",  7^7  and  70^ 

313.  There  are  two  consequences  of  the  last  two  principles,  of  con- 
siderable importance.  1st.  Whenever  it  is  required  to  extract  a  root 
of  a  complex  radical,  which  is  a  multiple  of  the  exponent  of  the  quan- 
tity under  the  sign,  the  extraction  can  be  indicated  by  suppressing  the 
exponent  under  the  sign,  and  multiplying  the  index  of  the  radical  by 
the  quotient,  arising  from  dividing  the  index  of  the  required  root  by 


252  FORMATION     OF    THE    POWERS    AND 

the  exponent  under  the  sign.  Thus,  let  it  be  required  to  take  the 
sixth  root  of  \/  «*,  the  result  will  he  y  a.     For,  the  sixth  root  of  v/  a? 

3    1  3  1  

is  equal  to  (a~)''  =  a      =  a"*  =  i/ a.     In  like  manner,  the  j^Jt""  root 

n     _1_  n  1 

of  iy"a"  =  ''v^-  For  the  j^n""  root  of  V a"  =  (a")?"  =  av^  =  aP"  = 
Va-  These  results  have  plainly  been  formed  in  accordance  with  the 
rule. 

314.  2d.  Whenever  it  is  required  to  raise  a  simple  radical  to  a  power 
which  is  a  factor  of  the  index  of  the  radical,  the  operation  can  be  per- 
formed by  dividing  the  index  of  the  radical  by  the  exponent  of  the 
power,  and  writing  the  quotient  as  the  index  of  the  radical  instead 
of  the  old  index.  Thus,  let  it  be  required  to  raise  Va  to  the  square 
power.     The  result  will  be  %/a;  for  the  second  power  of  the  sixth  root 

X  1        1  _ 

of  a  is  («^)^  =  a^^:^a^  =  V  a.     So,  likewise,  the  Ji""  power  of  '^ a 

\_  "J.  

z=  :^  a.     For  the  'n}^  power  of  V  •  a  =  (aP»)"  =  op"  =  ap  =-y  a. 

The  following  are  applications  of  the  consequences. 


EXAMPLES. 

1.  Required  d**"  root  of  -y/a*.  Ans.   i/a,  or  a. 

2.  Required  6*"  root  of  -y/a^.  Ans.   ^a. 

3.  Required  7«°  root  of  ^/a^'^.  Ans.  a. 

4.  Required  4*''  root  of  ^a^.  Ans.   %/ a. 

5.  Required  ^m*'' root  of  v^a".  Ans.  ^a. 

6.  Required  12'"  root  of  %/c^.  Ans.   l^oT 

7.  Required  4*''  power  of  i/a.  Ans.  a. 

8.  Required  «;*'>  power  of  t;/a.  Ans.  a. 

9.  Required  wm*"*  power  of  "^a.  Ans.   ^a,  or  a°. 

10.  Required  m*""  power  of  '"i/a.  Ans.  i/a. 

11.  Required  3*  power  of  lya.  Ans.  X/O" 


EXTRACTION    OF    ROOTS  —   ' 


12.  Required  12"'  power  of  ^a. 

13.  Required  6*"  power  of  v^a. 

14.  Required  S*"*  power  of  l^o. 

15.  Required  pni^^  power  of  '""^a. 

TO  MAKE  SURDS  RATIONAL  BY  MULTIPLICATION. 
CASE  I. 

Monomial  Surih. 

315.  Suppose  the  given  surd  is  ^  b  ;  this  is  equivalent  to  i°  .  Now, 

it  is  required  to  multiply  i^  by  such  a  quantity  as  will  make  it  rational. 
Since  the  surd  will  be  rational  when  the  numerator  of  the  fractional 
exponent  is  exactly  divisible  by  the  denominator  j  and,  since,  in  multi- 
plication, we  add  the  exponents  of  the  same  literal  factors,  the  multiplier 

of  J "   must  be  h,  affected  with  such  an  exponent,  that,  when  added  to 

— ,  the  sum  of  the  two  exponents  will  be  a  whole  number.  Call  x  the 
n 

unknown  exponent  of  the  multiplier,  then  x  -\ =  1,  or  .r  =  1 

11  11 

n  —  1  "-=-'  1         ^jil 

= .     Hence,  the  multiplier  is  i  "  ,  and  we  see  that  h''   X  h  " 

=  b,  &  rational  product.     Had  we  placed  x  -\ =2,  and  found  the 

multiplier  under  this  hypothesis,  the  product  would  have  been  h^. 
But,  when  it  is  required  to  make  the  surd  rational,  and  of  the  first 
degree,  the  sum  of  the  primitive  exponent,  and  the  unknown  exponent, 
must  be  placed  equal  to  unity. 

Let  it  be  required  to  find  a  multiplier  which  will  make  ?/■*   rational. 
Then,  X  +  i  =  l,  ov  x  —  i.     Hence,  the  multiplier  is  i/^,  and  we  see 

3         1 

that  ^^  .  1/'^  =y  is3L  rational  product. 
22 


254  FORMATION    OF    THE    POWERS    AND 


RULE. 

Place  the  primitive  exponent,  plus  x,  equal  to  unity  ;  find  the  value 
of  X  from  this  equation.  The  value  so  found  will  he  the  exponent  of 
the  midtiplier.  The  multiplier  itself  must  he  the  given  monomial,  ex- 
clusive of  its  exponent,  raised  to  a  power  indicated  hy  the  value  ofx. 


1.  Find  a  multiplier  which  will  make  x    -  rational. 

Ans.  X  "  . 

2.  Find  a  multiplier  that  will  make  y^  rational.  Aiis.  y". 


3.  Find  a  multiplier  that  will  make  y  °    rational.        Ans.  y' 

4.  Find  a  multiplier  that  will  make  y^+'"  rational. 

Ans.  y    ''-^'"     . 
Corollary.  ^ 

316.  The  principles  demontrated  in  Case  I.  enable  us  to  find  the 
approximate  value  of  fractions  whose  denominators  are  monomial  surds. 


RULE. 

Mxdtiply  both  terms  of  the  fraction  hy  such  a  quantity  as  loill  make 
the  denominator  rational.  Approximate  as  near  as  may  he  desired 
to  the  true  value  of  the  monomial  surd  in  the  numerator,  and  then  re- 
duce the  fraction  to  its  lowest  terms. 


5 
1.  Required  the  approximate  value  of     to  within  -01. 

Ans.  1*58. 
6  WW      5(3-16)  _  15-80  _ 


EXTKACTIOX    OF    ROOTS.  2c 

2.  Required  tte  approximate  value  of  — =  to  within  -001. 

Aj}s.  1-732. 

3.  Required  the  approximate  value  of  —=  to  within  -01. 

Ans.  2-52. 

4.  Required'the  approximate  value  of  — =:^=r  to  within  -001. 

VIOOO 

Ans.  3  162. 

6.  Required  the  approximate  value  of  —=  to  within   01 

Ans.  4-65. 

6.  Required  the  approximate  value  of  — z=:zr:  to  within  -0001. 

VIUUO 

Ans.  -189738,  nearly. 

7.  Required  the  approximate  value  of  — to  within  -1. 

;y4uo 

A71S.   54-2. 

8.  Required  the  appi'oximate  value  of     ___  to  within  -01. 

Ans.  2-32,  nearly. 

9.  Required  the  approximate  value  of  —z=  to  within  -000001. 

v/320 

Ans.   3-577781,  nearly. 

g 

10.  Required  the  approximate  value  of  — =iz  to  within  -001. 

v^2 

Ans.  2-519. 

15 

11.  Required  the  approximate  value  of  — =  to  within  -001. 

V  lo 

Ans.  3-873,  nearly. 

30 

12.  Required  the  approximate  value  of to  within  -001. 

Ans.  3-873,  nearly. 

13.  Find  the  approximate  value  of  —=  to  within  -01. 

Ans.  5  04,  nearly. 


256  FOR  31  ATI  ON     OF    THE    TOWERS    AND 


14.  Find  the  approximate  value  of  — ^=  to  within  -01. 

Ans.  -89. 

18 

15.  Find  tlie  approximate  value  of  — = 

^18'  Ans.  6-87. 


CASE  II. 

317.  To  find  a  multiplier  that  will  make  rational  an  expression,  con- 
sisting of  a  monomial  surd,  connected  with  rational  terms,  or  consisting 
of  two  monomial  surds. 

Let  it  be  required  to  make  rational  ^/p  ■}-  y/q  hj  multiplication. 
From  the  principle  demonstrated  in  Case  1.,  it  is  plain  that  y/jj  can 
only  be  made  rational  by  multiplying  it  by  v^^j,  and  V  q  can  only  be 
made  rational  by  multiplying  it  by  Vq-  But,  unless  v/p  and  <>/ q,  in 
the  multiplier,  are  connected  by  the  sign  minus,  there  will  be  two  terms 
in  the  product  remaining  irrational  and  unreduced.  Hence,  the  mul- 
tiplier must  have  the  minus  sign  between  its  terms. 

Thus, 


\/p  +  Vq 
Vp  —  Vq 

p-\-  s/pq 

—  Vpq- 

-9 

p  —  q 

If  the  given  expression  is-Jp —  Vq,  the  multiplier  must  be,  for  a 
like  reason,  Vp  -}-  V  q- 

If  the  given  expression  contain  but  one  monomial  surd,  and  is  of  the 
form  J)  -f  s/g-,  it  may  be  written  y/p^  +  Vq,  and  reduced,  as  before, 
by  multiplying  by  Vp^  —  V q-  So,  p  —  Vq  may  be  written  Vp''  — 
\/q,  and  may  be  made  rational  by  multiplying  it  by  Vj^"  +  y/q. 

It  matters  not  how  many  terms  may  be  under  the  radical,  or  how 
many  rational  terms  may  be  outside  of  the  radical,  the  foregoing  proces- 
ses will  still  be  applicable ;  because  the  sum  of  the  quantities  under  the 
sign  may  be  represented  by  a  single  letter,  and  the  sum  of  the  rational 
terms  outside  of  the  radical  may  be  represented  by  a  single  letter. 
Thus,  a  -\- b  +  -v/m  —  n  =  p  +  Vq  =  V p^  +  V q.  Multiply  now 
by  Vp^ —  Vq,  and  replace  p  and  q  by  their  values.  The  result  will 
be  (a  +  Vf  —  (w  —  ii). 


EXTRACTION    OF    ROOTS.  257 

Let  it  be  required  to  make  y/ p  +  v^g  rational  by  multiplication. 
The  operation  is  as  follows  : 

p  +  Vfq 

+  Vp'f-  +  ?  __        _ 
—  Vfq  —  Vp(l 

p  -j-  (J  =  Product. 

The  multiplier,  to  make  ^j^  rational,  must  be  ^jr  ;  and  the  multi- 
plier, to  make  ^q  rational,  must  be  ^  qK  But,  after  multiplication 
^y  Vl^  4-  ^^2^  there  remained  two  uncancelled  surds,  "</p^q  and 
V  q^p,  and  these  could  only  be  cancelled  by  multiplying  the  given  ex- 
pression by  —  Vpq- 

To  make  rational  ^'p —  ^  q,  the  multiplier  must  be  ^ p^  +  ^  (f 
-\-  \/pq.  The  multiplier,  then,  in  every  case,  is  the  sum  of  the  cube 
roots  of  the  squares  of  the  quantities  diminished  or  augmented  by  the 
cube  root  of  the  product  of  the  two  quantities,  according  as  the  sign 
between  the  surds  is  plus  or  minus. 

Let  it  be  required  to  make  Vp  -\-  >/ q  rational.  The  operation  is 
as  follows : 


4/p 

+  ^q 

■■  Given  surd. 

^p' 

-</q^- 

-  i/fq  +  '^pcf  = 

:  Multiplier. 

p  +  V?9 

-^pq'- 

-q 

+  </pV 

+  ^1^ 

li     —     ?     —         Product. 

Or  the  operation  may  be  performed  iu  this  manner : 

^P  +  v/  ?  =  Given  surd, 

Vp  —  y/  q  =  First  multiplier, 

VY—  v/5'=  /P—  /?=  First  Product, 

v'i>  +  -s/  q  =  Second  multiplier, 
ji    —    q=^  Second  product. 

Any  two  monomial  surds,  whose  common  index  is  some  power  of  2, 
may  be  reduced  in  the  same  manner.     And,  since  each  multiplication 


1258  FORMATION    OF    THE     POWERS    AND 

<;ive.s  a  product  containing  two  surds,  witli  a  common  index  one-lialf  as 
great  as  the  common  index  previous  to  multiplication,  it  is  evident 
that  the  number  of  multiplications  will  he  indicated  by  the  exponent 
of  the  power  of  2  in  the  primitive  index,  common  to  the  surds  given  to 
be  reduced. 

Thus,  %/j>  -}-  \/<2  can  be  reduced  by  three  multiplications,  since 
8  =  2'.  And  lyp  +  v/<?  can  be  reduced  by  four  multiplications, 
.since  IQ  =  2\  _  _ 

The  reduction  of  the  ^p  -f  ^q,  and  i^p  -f  [/q,  is  as  follows  : 

v/p  -r  \/'J  =  Given  surd, 

v/p  —  x/?  =  ^i^'^*-  multiplier, 

VF  —  V?  =  v/p"—  \/7=  First  product, 

i/jJ  +  V?  —  Second  multiplier, 

l/p^ —  X/<i=  s/p  —  Vl  —  Second  product, 
■s/p  +  \^9,  =  Third  multiplier, 
p  —      q  =  Third  product. 

V^  +  V'q=  Given  surd, 
\/p  —  v^2'  =  First  multiplier, 

V'f—V~q=  ^J—  ^^=  First  product, 

y/P  +  \^(l  =  Second  multiplier, 

x/p^ —  ^q^=^p  —  i/q  =  Second  product, 
i/¥+  V~q  =  Third  multiplier, 
v/p  —  y/q  =:  Third  product, 
Vp  +  \/<?  =  Fourth  multiplier, 
p    —    q  =  Fourth  product. 

We  have  this  process  for  the  reduction  of  -i/p  +  %/q. 

i/p  +  ^q  z=  Given  surd, 

s/yi  ^  VF—  W?  —  s^¥<f  +  n/p¥  =  Multiplier, 

p  +  ^yq 

+  ^pq'  +  ?  _  

—  -^J^'q  —  -^pY 

—  ^pY  —  ^pI        

+  j/pY  +  VpW 

p  +  q  =  Product. 


EXTRACTION     OF    ROOTS.  259 

The  v/p~+  \^2  can  be  reduced  in  a  similar  manner. 

\/p^+  %/(]_=■  Given  expression, 

fi/p  —  \/?  =  Fi'"'^^  multiplier, 

({/p^+  %/~^=  Vp  —  VT—  ^"'st  product, 

Vp^  —  V^  +  l/m  =  Second  multiplier, 
p  —     q  =  Second  product. 

All  expressions  composed  of  two  surds,  whose  common  index  is  a 
multiple  of  2  and  3,  may  be  reduced  in  a  similar  manner. 

Thus,  ^^p  +  '^q  =  Given  expression, 

y/p  —  v/g-  =  First  multiplier, 
Vi?  —  ^q  =  First  product, 
%/p  +  V?  =  Second  multiplier, 
^p  —  \/q  =  Second  product, 
Vp  —  Vq  +  Vpq  =  Third  multiplier, 
p  —      q  =  Third  product. 

All  expressions  composed  of  two  surds,  whose  common  index  is  some 
power  of  3,  may  be  reduced  in  the  same  manner  as  l/p  +  l/q. 
Take  as  an  example 

y/p  +  ^q  =  Given  expression, 

^P  +  V?—  Vpq=  First  multiplier, 

Vf  -f  V?  =  VF-^  yq~=  First  product, 

Vp^  +  s/q^  —  Vpq  =  Second  multiplier, 
p  +      5'  =  Second  product. 

Take  as  a  second  example,  %/p  +  %/q. 

Then,  %/p  +  %/q  =  Given  expression, 

%/f  +  ^?—  \/pq=  First  multiplier, 

X/F-^-  \/?^  =  Vp+  Vq  =  First  product,  which   can  be 

reduced  as  before. 

The  reduction  of  l/p  +1/  q,  is  more  difficult  than  any  of  the  pre- 
ceding? reductions. 


260  FORMATION     OF    THE    POWERS    AND 

We  will  write  the  terms  of  the  product  that  cancel  each  other  in  the 
same  vertical  column. 


1/ p  -'tl/q   =  Given  expression, 

^/+  77—  ^¥9—  ^p¥+  ^'pV+  l/YP—  v'W  =  Multiplier. 

P  +  VW9+1  +  v^—  VY^—  vW+  ^W+  W^ 
—  -y/?     —  ^^  +  '^p'^t + yv\^ — v^^'  —  ^-/vW 

p  +  2'  =  Product. 

All  expressions  composed  of  two  monomial  surds,  may  be  rendered 
rational  when  the  surds  have  a  common  index.  The  amount  of  diffi- 
culty attending  the  reduction  depends  altogether  upon  the  index. 
When  the  index  is  some  power  of  2,  the  reduction  is  very  easy.  But 
when  it  is  7,  11,  13,  17,  &c.,  the  reduction  is  difficult. 

When  an  expression  is  given  to  be  reduced,  we  must  first  examine 
the  factors  of  the  common  index,  and  make  our  reduction  correspond 
to  those  factors. 

Thus,  let  it  be  required  to  render  rational  l^p  +  l^g-.  The  factors 
of  15  are  3  and  5 ;  we  must  then  reduce  the  expression  by  the-  first 
multiplication,  so  that  the  common  index  of  the  result  shall  be  5. 

Thus,  v'p  +  v/?  =  Given  expression, 

15/p  _|_  y~f—  '4/pq=  First  multiplier, 

'5/pi~_|,  '.^^  =  ^p   -f-  ^q,  which  can  be  reduced  as  before. 

To  reduce  the  ^^/p+  ^^'q,  we  must  use  such  a  multiplier  as  will  leave 
a  common  index,  5,  in  the  product. 

Thus,  Vp   +  Vl, 


]yp^  —  10/^2  =  s/p  —  ^q^  which  can  be  made  rational  as 
before. 

Let  it  be  required  to  reduce  v'p  +  X/q. 

The  factors  of  the  index  are  3  (2)1  Therefore,  we  must  first  get 
rid  of  the  factor,  3,  and  reduce  the  expression  to  a  common  index, 
(2)'  or  8. 


EXTRACTION    OF    ROOTS.  261 

\/p  +  v/?  =  Given  expression, 
24/^  +  ^^J—'Vpq  =  First  multiplier, 
%/p   -{-  %/q  ^  First  product, 
Vp  —  \^9.    =  Second  multiplier, 
X/p  —  X/q  =  Second  product, 
Vp   +  v/?  ==  Third  multiplier, 
V'p  —  \/q  =  Third  product, 
yp   -\-  ^q   =  Fourth  multiplier, 
p  —      !/  =  Fourth  product, 

CoroUari/. 

318.  The  principles  developed  in  Case  II.  enable  us  to  find  the  ap- 
proximate value  of  a  fraction,  whose  denominator  consists  of  a  mono- 
mial surd  connected  with  known  terms,  or  of  two  monomial  surds. 

2 

Let  it  be  required  to  find  the  approximate  value  of  — =^ z:^ 

v/o  +  v/2. 

Th.n  ^  2(^5 -v/2)_  2(2-23 -1.41)       2(0-82) 

'*^'°' ^/5-+  ^-2  -         5-2         -  3  =-— -  = 

1-64 

— ^  to  within  -01. 

EXAMPLES. 


1.  Required  the  approximate  value  of  — = ~  to  within  -01 

Ans.  -59. 

2,  Requii-ed  the  approximate  value  of '■ — -  to  within  -01. 

^'"-  -59- 


For 


_3 ^3(8  — ^/5)^  3(8  —  2-23)      3(5-77)       17-31 

+-^5  64  —  5  59  69      ~~59~' 

3,  Required  the  approximate  value  of =:  to  within  -1. 

4+  V2 

Ans.  12-6 


262  FORMATION     OF    THE    POWERS    AND 


66     ^        66        ^     66  .(^(4/  +  V(2)^— ^(4)^  2) 
°''4+^"2~^4^+y'2     (^4='+^2X^(4f+  ^(27-^(4^) -2) 
_66  ((4^  +  1-6-5)  _^^^^ 
~  ^(4/ +  2  "    ■ 

2 

4.  Required  the  approximate  value  of  — = r^  to  within  -1. 

^4  +  ^2 

Ans.   -7. 

For  ^  _  HVW+  VW—  4/4^ _  2(2-5 +  1-6  — 2 

^1  +  y2  ~  4+2  6 

4-2 


5.  Required  the  approximate  value  of  =  to  within  -01. 

Ans.  13-23. 

6.  Required  the  approximate  value  of  — =: =:l  to  within  -01. 

Ans.  1-08. 

7.  Kequired  the  approximate  value  of     ^  to  within  -1. 

Ans.  14-4. 

30 

8.  Required  the  approximate  value  of  — n^  to  within  -1. 

V  25  +  V  5 

Ans.  6-4. 


9.  Required  the  approximate  value  of  -^rr to  within  -001. 

•s/6  —  \/5 

Ans.  37-44. 

2 

10.  Required  the  approximate  value  of  — r= =  to  within  -1. 

^^  ^4—^2 

Ans.  6-1. 


CASE  III. 

To  make  ratimial  an  expression   containing  three  or  more  terms 
of  the  square  root. 

319.  Let  it  be  required  to  make  rational  ^p  +  ^q  +  ^n. 

t 


EXTRACTION     OF    ROOTS. 

The  process  is  as  follows  : 

\/i>  +  y/q.    +  >/«  =  Given  expression, 
^/p  —  v/ J    +  x/«  =  First  multiplier, 

—  y/pi  —  ^/(^H  —  q^ 

p  —  q  -\-  11  -\-  'l^pn  =  First  product, 

p  —  q  -\-  n  —  2^pn  =  Second  multiplier, 

(P  —  ?  +  «)*  —  -^Pf-   =  Second  product. 

Let  it  be  required  to  make  rational  ^jj  +  v/'i  —  \/»- 
The  process  is  as  follows  : 

v^p  +  v/?    —  v/'i    =  Given  expression, 
\/i>  +  n/?    +  \/^i    ^^  First  multiplier, 

p  +  \^pq  —  s/pn 
+  Vpq  -\-     q  ~  >/q>^ 

+  s/pn  +  \/9»  —  n 


p  -{■  q  —  "  +  ~\/pq  =  First  product, 

p  -h  q  —  n  —  ~\/pq  =  Second  multiplier, 

(P  +  q  —  '0^  —  -^Pq  =  Second  product. 

Take  as  a  third  example  ^'p —  y/q  +  ^lii  +  v/«. 

^P  —  s^T  +  Vm    +  y/n 
y/p  +  >/q    —  ^m    +  ^n 


p  —  \/pq  +  -^pm  +  s/pn 
n  +  -^pn —  \/ nq  -\-   ^/  am 

—q  +   ^pq  +   \/nq  +  ^qm 

— m  —  s/jjni  —  y/mn  +  \/qm, 

(jj  +  %  —  q  —  m)  +  2-^pn  +  ^^  qm  =  First  product, 
(j3  +  «  — g  —  m)  —  I'sf  pn  +  2y/qm  =  Second  multiplier, 

P*       +  4P%/^TO  —  4j9tt  +  45'7?i  =  Second  product  reduced,  and 

in  which  P  represents 
the.  rational  term. 


264  FORMATION     OF    THE     POWERS    AND 

We  may  represent  P^  —  ij^n  +  4(pn  by  M^,  and  then  we  will  have 

M**  -f  AY^qm  =  Second  product, 
M^  —  4Pv^"^  =  Third  multiplier, 
M^  —  IQF'qm  =  Third  product. 


CoroUari/. 

320.  The  principles  of  Case  III.  enable  us  to  approximate  to  the 
/alue  of  a  fraction  containing  three  or  more  monomial  surds. 


EXAMPLES. 

1.  Reduce  — =: ^r^ to  an  equivalent  fraction  having  a  rational 

denominator.  Ans.  5+^/5 — v'30. 

For       _     ^^_  _  10(^5  —  VQ^+  1)  ^  ^ 

V5  +  V6  +  1     '   (V5  +  V0"+  1)  (v/T—  ^6"+  1) 
10(  s/5  —  V'G  +  1)       10  v'"5(  V  5"—  ^/ 0"+  1) 


2^5"  2(5) 


VSO  +  v/6. 


The  approximate  value  of  the  result  can  be  found,  if  desired. 

2 

2.  Reduce  —z= — ^rr.  to  an  equivalent  fraction  with  a  ra- 

V6+v/8+v/14 

tional  denominator.  A^is.  (\/6  —  \/8-f  V14(6  —  V 84) 

'  =48  ■ 

20 

3.  Reduce iz= :z= =  to  an  equivalent  fraction  with  a 

1+  V5+ v6+  v/10 
rational  denominator. 

Ans.   (1  —  V5  —  V6"+  n/TO~)  (v/TO  +  v/"30) 

__  _. 

Q 

4.  Reduce  — == = = — — =.  to  an  equivalent  fraction  with 

^1  +  ^/2  +  v/3  +  ^4      ^  ^ 

a  rational  denominator.  Ans.  (3  —  ^2  —  ^3)  (4  +  2^/6). 


EXTRACTION     OF     ROOTS.  265 

2 

5.  Reduce  — = = z= r=:to  an  equivalent  fraction  with 

^3  +  ^5  +  ^6  +  ^/8 

a  rational  denominator. 

Ans    (v/3"—  v/5"—  v/6  +  ^/8j  (^24  +  ^30) 
—  6 
2 

6.  Reduce  — = -= = =  to  an  equivalent  fraction. 

v/3 -f  v/5  +  >/6  +  v/10_  _  _ 

Ans.  v/3  —  v^S"—  v/(r+  v/TO. 


EXTRACTION  OF  THE  SQUARE  ROOT  OF  A  MONOMIAL  SURD 
CONNECTED  AVITU  A  RATIONAL  TERM,  OR  OF  TWO  MONO- 
MIAL SURDS. 

321.  Before  we  proceed  to  extract  the  root,  it  will  be  necessary  to 
demonstrate  three  principles  upon  which  the  extraction  depends. 

First  Principle. 

The  square  root  of  a  quantity  cannot  consist  of  the  sum  of  two  parts, 
one  of  which  is  rational  and  the  other  irrational. 

For,  if  possible,  suppose  \/ a  =  x  +  y/y.  Then,  by  squaring  both 
members,    we    get    a  =  x^  +  2xi/y  +  y-       From   which,    ^/y  = 

^ .     That  is  an  irrational  quantity  equal  to  a  rational  one, 

which  is  absurd.     But  the  absurdity  has  not  resulted  from  an  error  in 
the  analysis,  and  must,  therefore,  be  in  the  condition. 

Second  Principle. 

When  the  first  member  of  any  equation  contains  a  monomial  surd, 
connected  with  rational  terms,  and  the  second  member  is  made  up  in 
the  same  manner,  the  rational  quantities  in  the  first  member  are  equal 
to  the  rational  quantities  in  the  second,  and  the  irrational  quantities  in 
the  two  members  are  respectively  equal  also. 

Let  a  4-  \/b  =  X  +  i/i/,  then  a  =  x,  and  ■</ 6  =  -yy. 

For  if  a  be  not  equal  to  a;,  let  a  =  a:  ±  m.  Substitute  this  for  a 
in  the  equation,  and  there  results  x±m  +  v/^=  x  +  y/y,  or  ±  m 
-f  y/b  =  \/y,  which  is  impossible,  (principle  first.)  Therefore,  it  is 
absurd  to  suppose  that  a  is  unequal  to  x.  Hence,  a  =  x,  and  the 
equation  becomes  a  +  s/h  =  a  -\-  sf  y,  or  by  cancelling  a,  s/h  —  yjy. 
23 


266  FORMATION    OF    THE    TOWERS    AND 

Third  Principle. 


If 


\ /  «  +  \/^  =  a;  +  ^y,  then  will  y  /  a  —  ^/h  =  x  —  -^y. 
For,  by  squaring  the  first  equation,  we  get  a  -\-  ^/h  =  x^  -\-  2x\/y 

From  which  a  =  x^  +  y  (2),  and  ^h  =  2x^/2/  (3)- 

Subtract  (3)  from  (2),  and  we  have  a  —  ^Y=  x"  —  2x^y'+  y  (4). 

Extracting   the   square   root   of  both   members   of    (4),   we   have 

\  /  a  —  y/h  =  x  —  ■s/  y. 

This  last  equation  has  been  correctly  deduced  from  the  equation 
v/a  +  ^h  =  cc  +  y/y,  so  that  it  is  a  true  equation,  provided  that  the 
equation  from  which  it  is  deduced  is  true. 

322.  The  foregoing  principles  enable  us  to  deduce  a  formula  for  the 
extraction  of  the  square  root  of  an  expression  made  up  of  two  mono- 
mial surds,  or  of  one  monomial  surd  connected  with  rational  terms. 


Let,  \J  a  +  ^T=  X  +  ^y.     (1). 

Then,  \/  a  —  ^T=  x  —  ^Y-     (2)- 

Squaring  (1)  and  (2),  there  results 

a  +  Vh  =  x'  +  2x^yj-  y.     (3). 
a  —  ^h  =  x^  —  2xy/Y^y.     (4). 

Adding    (3)    and    (4),  and    cancelling    (2)    in  the  result,  we  get 
a=^x^  -^  y.     (5). 

Multiplying  (1)  and  (2),  we  get  ^a^  —  h  —  x'  —  y.     (6). 


Adding  (5)  and  (6),  we  get  "  "^  '^f ^  =  x\     (7). 


Subtracting  (6)  from  (5),  we  get "^ =  y.     (8). 

2 

Extracting  the  square  roots  of  both  members  of  (7)  and  (8),  and 
there  results  \ /^Liu^^^^ln^  _  x.     (9). 


EXTRACTION     OF    ROOTS. 


267 


and  s/'-^.f-'-Vj.     (10). 

By  adding  (9)  and  (10),  we  get  the  value  oi  x  +  y/y^ov  of  the 
equivalent,  \/«+  n/^- 

By  subtracting  (10)  from  (9),  tvc  get  the  value  of  x  —  ^i/,  or  the 
equivalent  \  /  a  —  \/^ 

Hence  we  have,  by  performing  these  operations,  the  two  formulaJ, 


(A). 


^/a  _  ^T=  \/»  +  ^^^-  \/"UZ^^I^.     (B). 

By  using  the  double  sign  =b,  we  may  unite  (A)  and  (B)  in  one 
formula. 


We  will  now  show  the  use  of  the  formula?. 


EXAMPLES. 

1.  Let  it  be  required  to  extract  the  square  root  of  9  +  v/45.  Then, 
by  comparing  9  +  ^/\:b  with  a  +  v/6,  the  quantity  whose  root  is  to 
be  extracted  in  (A),  we  see  that  a  =  9  and  h  =  45. 


Hence, 


s/a  +  ^T=  y/o  +  ^-35  =  y/S+^Sl-iS  _^ 


s/ 


9— V81  — 45 


v/71+  yl 


A  result  that  can  be  verified;  for,  squaring  it,  we  will  have,  H  + 

2.  Required  the  square  root  of  9  +  V72.     By  comparison,  we  get 
a  =  9  and  h  =  72. 


2CS  FORiNlATION     OF    THE    POWERS    AND 


And  from    (A)  we   get  v/o  +  V72  =  \ /^  +  V81  — 72  _^ 

This  result  can  be  verified;  for,  squaring  it,  we  get  6  +  2^18  +  3 
=  9+  v/72: 

3.   Required  the  square  root  of  9  —  v^56. 

Then,  a  =  9  and  h  =  56.     Formula  (B)  must  be  used. 


From    (B)    we   get    \J  ^  —  ^/56   =   \/^  +  ^^^  ~  ^^  _ 

.  /9  —  ^8r=r5B 

V  2 =  \/7  —  n/2. 

Squaring  this  root,  we  have,  7  —  2v'14  +  2  =  9  —  v/56, 
4.  Required  the  square  root  of  9  +  2%/  —  162. 


Pass  the  2  under  the  radical,  thus,  9  +  2^— 162  =  9+  V—  648, 
and  v/fl^  — &=  V81  +  648  =  V  729  =27,  and  the  application  of 

the  formula  will  give  \ /9  +  2v/'=^62  ==  Vl8+  V— "IT. 

This  result,  when  squared,  will  give  18  -f-  2  s/ — 162  —  9  =  9  + 
2v/ —  162.     Hence,  the  result  is  true. 


5.  Required  the  sum  of  \/9  +  2v^— 162,  and  \J^—2-/—lm. 

Ans.  2VT8: 

This  result  can  be  verified ;  for,  2  V 18  squared  =  72 ;  and  the  given 
expression,  when  squared,  produces  9  +  2s/ —  162  +  9  —  2v/ —  162 
+  2s/81  +  648  =  18  +  2v/729~=18  +  54  =  72. 

6.  Required  the  square  root  of  \/8  +  ^/ —  1. 


A^.  ^^^  +  ^^^ 


2         ^   V  2 

This  result  can  be  verified  by  squaring  it.     We  will  get  -f- 


n/8  — 3       ^ 


/8  — 9         ,_        ^ 


EXTRACTION    OF    ROOTS.  269 

7.  Required  the  square  root  of  1  +  V —  15. 

Ans.   VT+  v/^. 
Verify  tte  result  by  squaring  it. 


8.  Required  the  square  root  of  1  +  2>/ —  20. 

^ns.   V^  +  v/ — 4, 

Verify  the  result  by  squaring  it. 


9.  Required  the  square  root  of  1  +  V — 288. 

Ans.  3  +  ■/■ 
Verify  the  result  by  squaring  it. 


10.  Required  the  sum  of  \/l+s/—2W,  and  \/l— v/^2BB. 

Ans.  6. 

We  see  that  the  square  root  of  an  expression  containing  an  imagi- 
nary quantity,  cannot  have  all  its  terms  real.  But  the  sum  of  the  roots 
of  two  expressions,  involving  imaginary  quantities,  may  be  real  and  even 
rational.  The  formulas  (A)  and  (B^t  are  generally  applied  to  expres- 
sions for  which  y/ a" — b  is  rational. 

11.  Reduce  — = to  an  equivalent  fraction  with  a  rational 

V9  -I-  V*i  _ 

denominator.  Ans.  10  (y/o  —  v/2). 

For        ^^       =  10  ( vw  -  vw  -  Vim+i/Wf)  ^ 

=  2  (v/27  —  ^Y—  ^/T8'+v/T2)=2(3v/'3  —  2 ^2"— 3 ^2+ 2^/8) 
=  2  (5v/3"_  5^/2)  =  10  (v/3~—  ^2). 

Or,  the  reduction  may  be  performed  thus,  -^— — -  =  — ^ 

(^3-fv^2)(^3_^2  3-2  ^^  ^    ^ 

4 

12.  Reduce  — r= =  to  an  equivalent  fraction  with  a  rational 

V5-ft/4  '         _      _  _ 

denominator.  Ans.  4  (  (2  +  v/o)(n/5)— 2^/2  —  y/KJ). 

23* 


270  FORMATION    OP    THE    POWERS    AND 


4  _      ( yy  —  4/(4)^  —  V(5/4  +  V5(4)^)  _ 

_V5+Vi~_  _  5-4     _ 

4(^5  4/5  — v/8  —  v/l<^  + 2^5)  =  4((2+^5)(^5)_2^2  — 

n/10). 

5 

13.  Reduce  — = ^^  to  an   equivalent  fraction  with  a  rational 

y4  +  i/2  _  _ 

denominator.  Ans.   |  (^4  -(-  (3/2  —  1)2). 


p^^       _5  ^  ^  (y  (4)^  +  V(2)^  _  3/(4)  (2) 


y4  +  3/2_        (y4  +  y2)  (^(4)^  +  V(2)^  -  VC4)  (2)  ) 
=  I  (2^2  +  3/4  _  2)  =  1(3/4  +  (^2  _  1)2). 

14.  Reduce  — t=z= —  to  an  equivalent  fraction  witli  a  rational 

y27+V8 
denominator.  Ans.   ^. 


For       _^  =   6  (^(27)^+  V(8r-y(27)  (8)) 

'   V27  +  y8  35 


The   result  is   rational   because   the   given   expression   was    really 
rational,  though  under  an  irrational  form. 

15.  Reduce  :=^  to  an  equivalent  fraction  with  a  rational  de- 

n  -f-  V'wi  

nominator.  .        m  (n  —  \/m) 

Ans.    3 . 


IMAGINARY  QUANTITIES. 

323.  Any  expression  whatever,  made  up  of  monomial  surds  and 
rational  terms,  or  monomial  surds  only,  may  be  rendered  rational  by 
repeated  multiplications.  The  few  examples  given  will  show  the  man- 
ner in  which  these  multiplications  must  be  made.  No  general  rule  can 
be  given  in  regard  to  them.  There  is  a  particular  class  of  monomial 
surds,  which,  when  rendered  rational,  present  some  differences  from 
other  surds  in  regard  to  their  algebraic  signs.  These  are  imaginary 
surds. 

An  imaginary  quantity  has  been  defined  to  be  the  even  root  of  a 
negative  quantity,  because  no  quantity,  taken  as  a  factor  an  even  num- 
ber of  times,  can  give  a  negative  result.  Thus,  y/ —  a,  %/ —  o, 
%/ —  o,  ^;/ —  a,  are  imaginary  quantities. 


EXTRACTION    OF    ROOTS.  271 

If  tlie  indicated  root  be  of  the  2ni"'  degree,  then  ^^ —  a  may  be 
written  ^^a  ^^ —  1  =  6\/ —  1.  In  which,  h  represents  the  27H*'' 
root  of  a,  whether  that  be  rational  or  irrational.  And  we  see  that  all 
imaginary  quantities  may  be  decomposed  into  two  factors,  one  of  which 
is  an  indicated  even  root  of  minus  unity,  and  the  other  of  which  is 
real,  and  sometimes  rational. 

This  decomposition  must  always  be  first  effected  previous  to  opera- 
ting upon  imaginary  quantities. 

To  square  v/ —  «;  "we  first  write  the  expression  v/o  v/ —  1 ;  these 
two  factors  will  both  drop  the  radical  when  squared.  Hence,  (^Z —  a)^ 
=  W^  V-^^=  («)  (-  !)=—«. 


So,  also,  (y—of  =  (v/a  v/— 1)'  =  (ds/a)  (—1  y/—l)  = 
Qx/ — a- 


(^— «)^  =  (v/<(  ^/—  l)'  =  Or)  (+  1)  =  +  a' 


(v/—  af=(ya  ^—  iy=(a'  ^a)  (+^/—  l)  =  aV- 


(v/— a/=(v/a  v/-  !/=(«')  (-  1)  =  —  «'• 

The  table  shows  that  tlie  even  powers  of  imaginary  monomials  are 
always  real,  and  that  their  signs  are  alternately  plus  and  minus. 

324.  A  table  of  products  will  show  the  modifications  of  their  alge- 
braic signs. 


(  +  x/— ")(  +  n/— »):^+(v/av/— l)(v/av/— 1)  =  (^a) 

(+v/— «)  (— y/— g)^  (^^^ZTT)  (_^^^^Q  =  (y-^ 
C-v/«)(n/-1)(n/— !)  =  +«. 


C+yZ—g-)  (+v/— ^)  =  i^a  ^—  1)   {^b  y/—  1)   =   (^a) 
(x/i)  (>/—  1)  (v/=l)  =  -x/^ 


(— n/—  «)  (— %/—  6)=(— v/a  y/—  1)  (— v/6  y/—  1)=(— ^a) 

(-V/^  (y/=^  (x/^^)  =  — x/"^ 


272  ro  UMAX  ION  or   the   powers  and 


(-fV-<0  (— v/— /^)  =  (v/a  V— 1)  i—s/h  V— 1)  =  (v/a) 
(— v/TT)  (v/^^1)  (v/— 1)=+^"^. 

We  see  that  like  signs  produce   minus,  and  unlike  signs  produce 
plus,  wlien  imaginary  monomials  are  multiplied  together. 

325.  "We  will  now  form  a  table  of  quotients. 


^■v- 

—  a 

_  v/av/— 1  _ 
v/6v/— 1 

+  v/« 

+  v/- 

+  v/6 

+  v/— 

a 
1" 

-v/^x/— 1 

v/^ 

— v/— a 
-v/-6 

=   - 

— v/T  v/^1         - 

-v/i 

v/a 

— V- 

-a 

—v/a  v/—  1 

-v/TT 

+>/"= 

=^' 

v/6  v/— 1 

v/6 

And  we  see  that,  in  division,  like  signs  produce  plus,  and  unlike 
signs  produce  minus.  It  would  seem  that,  since  the  rule  for  the 
signs  in  division  is  different  from  that  in  multiplication,  the  product 
of  the  quotient  by  the  divisor  might  not  give  the  dividend.  But  any 
of  the  preceding  quotients  can  be  readily  verified. 

Take^^=±^ 

V—h        +v/i        _ 

+  -y/g  _  _  4-  v/a 

Multiply  the  quotient  ~j  by  the  divisor  v/ —  h.  Then,  — — -= 

+  v/^ 


X  v/ —  b  = ^  y/b  v/ —  1  =  -f  \/a  >/ —  1  =  \/ —  a,    the 

dividend. 

326.  Imaginary  quantities  can  be  operated  upon  just  as  real  quan- 
tities, provided  that  care  be  given  to  attribute  the  proper  algebraic  signs 
to  the  results. 

Let  it  be  required  to  divide  4v/ —  i*  by  2by/ —  1.  The  division 
cannot  be  performed  until  4v/ —  i^  is  transformed  into  the  equivalent 
expression,  4&v/ —  1?  tlic  quotient  will  then  be  2. 

Let  it  be  required  to  multiply  4v/ —  b^  by  2&v/ —  1-  Then, 
4v/^T''  X  2b^'^l  =  Uy^l  X  2b^'^^=—Sb^ 


EXTRACTION     OF    ROOTS.  273 

Now  let   it   be   required   to   divide   — 86^   by   2.h^ —  1.      Then, 
—86^  SiV^^^v/"^^^^  ^  _  4^  xTZT: 


1h^—\  26v/— 1 

EXAMPLES. 

1.  Multiply  +  x/ —  a  by  —  ^ —  h.  Ans.   +  y/ah. 


2.  Divide  ^ab  by  x/ —  a.  Ans.  —  ^ — 6. 

y/ab    _     s/a  y/h     _     y/ZT    _  —  v/6^n/—  Iv/—  1  _ 

—  n/6'n/— 1  =  —  V^^>. 

The  minus  sign  is  placed  before  the  square  root  of  h,  in  the  expres- 
sion, —  v/6  v/ —  1  v^ —  1,  in  order  to  make  the  result  positive. 
Since  ^ — 1  y/ — 1  =  —1. 

3.  Divide  v'"*^  by — V — ^-  ^"S-   +  v^ — <*• 

4.  Multiply  ai^ —  c"  by  •/ —  a^bc.  Ans.  —  a^bc  ^bc. 


5.  Divide  —  a^bcy/bc  by  y/ —  a-bc.  Ans.    -f-  abc  y/ —  1. 

6.  Divide  —  a%Cy/bc  by  (fi^/ —  ^^-  '4?is.    +  «  s/ —  be. 


7.  Multiply  —  V—  abc  by  —  abc y/— abc.  Ans.  —  a?b^<?. 


Divide' — a^6V  by  —  v/ —  aic.  Ans.  —  abc  V —  abc. 


9.  Divide  —  a^Z*V  by  —  abc  yf —  abc.  Ans.  —  V —  abc. 

10.  Add  together  asj —  6-  and  ^v/ —  a^.  Ans.  2.ab-s/^^. 

11.  Eequired  the  square  power  of  a  +  hsj —  1. 

Ans.  a^  +  2.abs/—\  —  h"^. 

12.  Required  the  square  root  of  a^  +  2a5-/— "l  —  ^<^ 

-4ns.  a  -f  i'/ —  1- 

13.  Required  the  third  power  of  a  +  by/ —  1. 

Ans.  a'  —  ZaV  +  Sa^^^v/^a  —  ^V'^T. 


274  FORMATION    OF    THE    POWERS    AND 

14.  Required  the  product  o{  a  4-  hs/ —  1  and  a  —  h^ —  1. 

Ans.  a"  +  y. 

15.  Required  the  quotient  of  a^  +  Z*^  by  a  +  hy/ —  1. 

Ans.   a  —  hs/ — 1. 

16.  Required  tte  quotient  of  a^  +  Why  a  —  hs/ —  1- 

Ans.   a  +  h\/ —  1. 

17.  Required  the  cube  root  of  ci"  +  Za^hJ^-l  —Sah'  —  h\/^. 

Ans.   a  -\-  hs/  —  1. 

18.  Required  the  fourth  power  of  a  +  &>/ —  1. 

Ans.  a"  +  '^aj'h  JZI\  —  QaW  —  ^aV V— 1  +  h\ 

19.  Subtract  2  2a^  from  10vA=^.  Ans.  8v/^=^. 

20.  Add  together  8x7"=^  and  22/Zr2.  Ans.  lOs/— 2. 


21.  Muhiply  V— a  +  -J—h  by  v  — «—  %/— 6. 

Ans.  h  —  a. 

22.  Divide  h  —  «  by  v' — ^  +  7 —  a. 

Ans.  —  ■/ — h  +  V — a. 

23.  Multiply  J-  +  2  s/^=r^  by  J  —  2  ^—a.  Ans.   \  +  4a. 

24.  Divide  I  +  4a  by  i  +  2v7Zr^.  ^^s.  i  —  2x/'^^.. 

25.  Multiply  2  t/^Ta  by  3  v/^T^.  ^ns.  6V^^  V— ^ 

26.  Divide  6v/— a  ^  —  a  by  2^^^^.  ^hs.  3s/-^^. 

27.  Divide  6v/—  a  V^=^  by  3  v/^^f~  ^hs.  2y^^. 

28.  Multiply  V^^^+  t/^^6'by  y^=^—  y"^^=^. 

J.J2S.    V — a — v/ — i. 

29.  Multiply  1  +  6v/'^=T+  cV^irTby  l_Z)V^^^nr+  cV^lT 

^}»S.     1    +  ^2  _  g2  ^   2cn/—  1. 

80.  Multiply  1  +  6v''=T  +  c^/— "l  by  1  +  Z/^I^  —  Cy/^^TT" 


EXTRACTION    OF    ROOTS. 


31.  Multiply  5v/—  1  +  V—  3  by  3v/—  1  —  V—  ^7- 

Ans.  —15  —  3v/3'+  15^/3~+  9. 

32.  Required  the  second  power  of  5v/ —  1  +  ^Z  —  3. 

Ans.  — 25 — lOv/o — 3. 

33.  Required  the  square  root  of  —  25  — 10^/3"— 3. 

Ans.  5v/IIT— 3. 

34.  Divide  6  4-  v/ — 4  by  2  +  y/ — 9,  and  make  the  denominator 

of  the  quotient  rational.  ^        18  —  14*/ —  1 

Ans.  -^3 . 


For     ^+  ^-j    =    (6+  v/-4)     (2  -  y/  -  9) 
2  +  v/— 9  (2+s/— 9)  (2  —  ^—9) 

12     +     2^—   4    —    6^—   9—    y/    3(j    y/—    1     y/—    1  _ 

___  4  +  9 

12  +  4v/— 1  — 18y/"=nr+  6  _  1«  —  14v/^^ 
13  ~  13  ■ 

35.  Divide  2  +  V — 2  by  2 —  v^ — 2,  and  make  the  denominator 

of  the  quotient  rational.            ,          (2  +  y/—  2)^       2  +  4y/ir2 
^ns.     -^^ -^ or 


G  (5 

30.  Divide  1  -|-  y/ —  1  by  1  —  ■/ —  1,  and  make  the  denominator 
of  the  quotient  rational.  Ans.   y/ — 1. 

37.  Required  the  quotient  of  Gy/ — 4  divided  by  2y/ — ~^. 

Ans.  2. 

Rcmarhs. 

An  examination  of  the  foregoing  results  will  show  some  properties 
of  imaginary  quantities. 

1.  Real,  and  even  rational  results,  may  be  obtained  by  the  ordinary 
algebraic  operations  upon  imaginary  quantities. 

2.  Two  monomial  imaginaries  raised  to  an  even  power,  will  always 
give  a  rational  result. 

3.  Two  monomial  imaginaries  multiplied  together,  or  divided,  the  one 
by  the  other,  will  always  give  a  real  result. 

4.  A  monomial  imaginary  connected  by  the  sign,  plus  or  minus,  with 
one  or  more  rational  terms,  and  then  raised  to  any  power,  will  give  at 
least  one  imaginary  term  in  the  result. 

5.  Any  expression  involving  one  or  more  monomial  imaginaries,  may 
be  rendered  real  by  one  or  more  multiplications. 


276  EQUATIONS     OF     THE     SECOND     DEGREE. 


EQUATIONS  OF  THE  SECOND  DEGREE. 

827.  An  equation,  involving  a  single  unknown  quantity,  is  said  to 
be  of  the  second  degree,  wlicn  the  highest  exponent  of  that  unknown 
quantity  in  any  one  term  is  2.  Thus,  x^  -\-  x  =  m,  and  x^  -{-  a  =  h, 
are  equations  of  the  second  degree  with  one  unknown  quantity. 

An  equation,  involving  two  unknown  quantities,  is  said  to  be  of  the 
second  degree  when  the  highest  sum  of  the  exponents  of  the  unknown 
quantities  in  any  term  is  equal  to  2.  Thus,  xt/  =  ?h,  a;^  -f  ?/  =  ??,  and 
y'^  -{-  x=p  are  equations  of  the  second  degree  with  two  unknown 
quantities.  For  they  may  be  written,  x'j/'  =  in,  x^if  -f  ^  ^=  n,  and, 
y^x^  -f  a-  =  w,  and,  thei'efore,  come  within  the  definition. 

We  will  begin  with  equations  of  the  second  degree,  with  a  single 
unknown  quantity.  These  are  divided  into  two  classes,  complete  and 
incomplete.  A  complete  equation  of  the  second  degree,  with  a  single 
unknown  quantity,  is  one  which  contains  the  second  and  first  powers 
of  the  unknown  quantity.  Thus,  a-^  +  a  =  0,  clx^  +  x  =  m,  and 
ax'^  -f  hx  =  c,  are  complete  equations  of  the  second  degree,  with  one 
unknown  quantity.  There  may  or  may  not  be  other  terms,  besides 
those  involving  the  unknown  quantity.  The  first  and  second  powers 
of  the  unknown  quantity  may  or  may  not  have  coefficients  different 
from  unity. 

An  incomplete  equation  of  the  second  degree  is  one  containing  the 
second  power  only  of  the  unknown  quantity,  and  it  may  or  may  not 
involve  terms  in  which  the  unknown  quantity  does  not  enter.  The 
unknown  quantity  may  or  may  not  be  affected  with  a  coefiicient  diffe- 

rent   from    unity.     Thus,  ax^  =  x^,  x^  -\-  m  ==  n,  ~ — h  c  =  —  q  are 

incomplete  equations  of  the  second  degree.  Equations  of  the  second 
degree,  with  one  unknown  quantity,  are  frequently  called  quadratic 
equations,  because  the  unknown  quantity  is  raised  to  the  second 
degree,  or  squared. 

INCOMPLETE  EQUATIONS. 

328.  We  will  first  examine  incomplete  equations. 

hx^ 
The  general  form  of  such   equations   is    ax^ \-  d  =^e.     By 

transposition   and   reduction,  we  get  (ca  —  h')x'^  =  ce  —  cd,  or   c'^  = 


EQUATIONS  OF  THE  SECOND  DEGREE.        2i( 

~ -^  =  q,  by  substitutinsr  for  the  known  terms  in   the  second 

ca  —  6 

member  a  single  term,  equal  to  them  in  value.  Hence,  ever}-  incom- 
plete equation  may  be  reduced  to  two  terms,  and  the  equation  be 
placed  under  the  form  of  x^  =  q.  Owing  to  this  circumstance,  incom- 
plete equations  are  sometimes  placed  in  the  class  of  hinomial  equations, 
or  equations  involving  but  two  terms. 

There  is  no  difficulty  in  solving  the  equation  x^  =  q.  The  square 
root  of  the  first  member  will  give  us  x  ;  but  if  we  extract  the  square 
root  of  the  first  member,  we  must  extract  that  of  the  second  also,  or 
the  equality  will  be  destroyed.  Extracting,  then,  the  root  of  both 
members,  we  have  dti  x  =  zt:  ^/  q.  We  have  prefixed  the  double  sign 
to  both  members,  because  this  equation,  when  squared,  must  give  back 
the  original  equation,  x^  =  q,  from  which  it  was  derived,  and  either 
+  X  or  —  X,  when  squared,  will  give  +  x'^,  and  either  +  V  q,  or 
—  ^/q  will,  when  squared,  give  q. 

The  value  of  the  unknown  quantity,  or  the  root  of  the  equation,  as 
it  is  generally  called,  has  been  defined  to  be  that  which,  substituted 
for  the  unknown  quantity,  will  satisfy  the  equation,  that  is,  make  the 
two  members  equal  to  each  other.  We  sec  that  an  incomplete  equa- 
tion of  the  second  degree  has  two  values,  numerically  ccjual,  but 
aff"ectcd  with  different  signs.  Either  value  or  root  will  satisfy  the 
equation,  for  the  result  of  the  substitution  of  either  V*/,  or  — y/q  for 
X,  in  the  given  equation,  will  he  q  z=q.  We  have  prefixed  the  double 
sign  to  both  members,  but  it  is  usual  to  prefix  it  only  to  the  root.  In 
that  cape,  however,  we  must  understand  that  the  sign  of  x  is  not  neces- 
sarily positive,  it  being  afi'ected  with  the  positive  sign  only,  when  it 
corresponds  to  the  positive  root.  It  becomes,  then,  necessary  to  have 
some  notation  to  distinguish  the  values.  This  we  will  do  by  dashes. 
The  X  that  corresponds  to  the  positive  root  we  will  write  x' ;  and  the 
X  that  corresponds  to  the  negative  root  we  will  write  x".  The  fir.st  is 
read  x  prime,  and  the  second,  x  second.  It  will  be  shown  that  a  com- 
plete equation  has  two  roots,  generally  unequal  in  value.  We  will 
also  distinguish  these  roots  or  values  by  writing  them  x',  and  x".  Some 
equations  of  high  degrees  have  three,  four,  &c.,  values.  These  will  be 
written  x',  x",  x",  x^'',  &c.,  and  read  x  prime,  x  second,  x  third,  x 
fourth,  and  so  on. 

To  solve  an  incomplete  equation,  it  must  first  be  put  under  the  form 
of  x'^  =  q.  That  is,  all  the  known  terms  must  be  transferred  to  the 
second  member,  and  the  coefficient  of  x"^  must  be  made  pins  unity,  if 
24 


278       EQUATIONS  OF  THE  SECOND  DEGREE. 

not  already  so,  by  dividing  both  members  by  the  coefficient  of  x-. 
After  the  equation  has  been  put  under  the  form  of  x"  =  q,  extract 
the  square  root  of  both  members. 


EXAMPLES. 

1.  Solve  the  equation,  2x^  +  1=4. 

Transposing,  and  dividing  by  the  coefficient  of  x"^,  we  get  x^  =  ^. 
Hence,  x  =  ±  vT.  Then,  x'  =+y/J,  and  x"  =  —Vj.  The  solu- 
tions may  be  left  thus,  or  we  may  extract  the  indicated  root  approxi- 
matively. 

2.  Solve  the  equation,  2x^  +  4  =  1. 

Reducing,  we  get  x^=±^ — ^.      Then,  x' = -\- y/ — ^,  and  x" 

These  roots  are  imaginary.  How  are  we  to  interpret  them  ?  An 
imaginary  quantity  indicates  an  impossible  operation.  Ought  not  the 
equation  which  produces  it  involve  an  impossibility  ?  In  this  instance, 
2x'^,  an  essentially  positive  quantity,  is  added  to  4,  and  their  sum  is 
required  to  be  less  than  4.  The  condition  of  the  problem  is,  then, 
absurd,  or  impossible,  and  the  result  is  impossible,  as  it  ought  to  be. 
It  will  be  shown  more  rigorously,  hereafter,  that  an  imaginary  solution 
always  indicates  absurdity  in  the  conditions  of  the  problem.  The  ima-/ 
ginary  values,  though  they  fail  to  satisfy  the  conditions  of  the  problem, 
yet  will  satisfy  the  equation  of  the  problem ;  as  they  manifestly  ought 
to  do,  since  they  have  been  truly  derived  from  it.  Substituting  either 
+  v/ — ?>,  or  — v^ — -?,  for  x,  in  the  equation,  2.x^  +  4  =  l,we  get  —  H 
+  4  =  i,  or  1  =  1.    ~ 

329.  We  observe,  then,  this  remarkable  analogy  between  imaginary 
values  in  equations  of  the  second  degree,  and  negative  values  in  equa- 
tions of  the  first  degree.  The  values,  in  both  cases,  will  satisfy  the 
equation  of  the  problem,  but  will  not  satisfy  the  required  conditions. 
There  is,  however,  this  difi"erence :  to  convert  a  negative  solution  into 
a  positive  one,  numerically  equal  to  it,  we  have  only  to  impose  a  single 
condition;  but,  to  convert  an  imaginary  solution  into  a  real  solution, 
precisely  equal  to  it  numerically,  we  have  generally  to  impose  two 
conditions.  Thus,  the  equation  which  gives  the  imaginary  solutions, 
expressed  in  words  would  read  thus  :  required  to  find  a  number,  twice 
the  square  of  which,  augmented  by  4,  will   give  a  sum  equal   to  1. 


EQUATIONS  OF  THE  SECOND  DEGREE.       279 

The  values  are  cc' =  +  ■«/ — 3,  and  a;"  =  — \/ — |.  Now,  to  get 
real  values  numerically  equal,  and  not  to  make  any  change  upon  the 
arithmetical  values  of  the  known  quantities  in  the  equation,  it  must 
be  changed  into  2aj^  —  4  =  —  1.  This  equation,  expressed  in  words, 
would  read  thus :  required  to  find  a  number,  twice  the  square  of 
which,  diminished  by  4,  will  give  a  difference  equal  to  —  1.  The 
values  of  the  new  equation  are  cc' =  +  \/ -|- |,  and  x"  =  — V+l- 
These  values  are  real,  and  differ  from  those  in  the  first  equation  only 
in  the  signs  of  the  quantities  under  the  radicals.  To  get  these  new 
values,  we  made  no  change,  arithmetically,  upon  the  numbers  in  the 
first  equation,  but  we  made  two  changes  of  sign,  or,  in  other  words,  we 
imposed  two  conditions  upon  the  equation  of  the  problem. 

3.  Solve  the  equation,  x^  +  h  =  a. 


Ans.  x'=z-^\/a  —  bj  x"  z=z — s/ <-''  —  h. 

Now,  let  a  =4:,  and  6  =  6.  Then,  a;'  =  -f  1/ —  2,  and  x"  =  — 
y/ — 2.  To  get  the  real  values,  x' =  +  >/ 2,  and  x"  =  —  %/2,  we 
must  make  h  and  a  interchange  values,  b  must  be  4,  and  a,  G ;  that 
is,  two  conditions  must  be  imposed. 

4.  Solve  the  equation,  ^  —  1  +  x"  —  4  +  5  =  3x='  —  7. 


Ans.  x'  =  +2,  x"  =  —  2. 


5.  Solve  the  equation,  -  -  _  1  -|-  .^2  _  9  _,_  5  =  3_,.2  _  22. 

Ans.  cc'  =:  +3,  x"  =  —  3. 

x^ 

6.  Solve  the  equation,  —  —  1  +  .x^  _  IC  +  5  =  Sx^  —  43. 

Ans.  a;'  =  +  4,  x"  =  —  4. 

7.  Solve  the  equation,  'jx^ —^  +  4x'  —  20  +  IOO.t;^'  —  500  = 

o 

995  —  199x1  Ans.  x'  =  +  v'S^  x"  =  —^'b. 

8.  Solve  the  equation,  ^+  ^  _|._ 3+ 7^2— 42  +  999x2==  5994. 

Ans.  x'  =  +  ^6,  x"  =  —  v^ey 

The  preceding  examples  are  simple,  and  the  solutions  can  be  readily 
obtained.  But  there  are  many  of  a  more  complicated  character,  and 
of  course  more  difficult  to  solve.  No  general  rules  can  be  given  to  aid 
the  student  3  he  must  exercise  his  own  ingenuity.     It  is  well,  however, 


280  EQUATIONS    OF    THE    SECOND    DEGREE. 

to  make  the  clearing  from  fractions  the  firet  step  in  every  reduction, 
and  then,  if  there  is  a  single  radical  in  the  equation,  it  ought  to  be 
placed  in  one  member  by  itself. 


1.  Solve  the  equation, ^  3  =  -r 

'44 


Clearing  of  fractions,  we  get  ^/x^  +  144  +  12  =  x^. 


Placing  the  radical  by  itself,  we  have  y/x"^  +  144  =  x^  — 12. 
And  squaring  both  members,  there  results  x^  +  144  ^=  x'^  —  24x'^  + 
144.     Hence,  a"  =  ^bx",  or  x^  =  25.     Then,  x'=  +  5  and  x"  =  —  5. 

x  10 


2.   Solve  the  equation, 


+  ^x'  +  44       22- 


Then,  22a;  =  lOx  +  10^/x^  +  44,  or  12x  =  IQ  ^  x^  +  44. 
Squaring  both  members,  we  get  144^^^  =  lOOx^  +  4400,  or  44a;'' 
4400.     Hence,  x'  =  10  and  cc"  =  —  10. 

^7'2~Iir77v 
3.  Solve  the  equation,  — '■ — ^ +  Z>  =  x^ 


Ans.  x  =  — — — ,  X  ■  =  —  — . 

c  c 

These  results  can  be  verified  by  substituting  either  the  value  of  x'  or 

x"  for  X  in  the  given  equation.     Then  we  will  have  \  / ^ \-  V(? 

A  +  2lc^       A          ,              .       ,     ,            ,         1  +  Sic'^       ,,, 
=  cl -^ 61,  or,  by  squaring  both  members, ^ 1-  6-r 

Hence  the  equation  is  satisfied. 

4.  Solve  the  equation, ;==r  =  — . 

X  +  ^x^  -\-a        a 

+  &  „  -6 


^??S.    X'  := 


-v/a  —  26  v/^f 


EQUATIONS    OF    THE    SECOND    DEGREE.  281 

Verify  these  results.     What  do  they  become  when  a  =2b?     "Why  ? 
What,  when  a  =  0  ?     Why  ?     What,  when  26  >  a  ?     Why  ? 


5.  Solve  the  equation,  ^p^  +  x^  +  x  =  mjc. 


Av    -'  —  '  ^         -"  —  ^ 


ins.  X  = 


Verify  these  results.  What  do  they  become  when  m  =  2  ?  ^^hy  ? 
What,  when  wi  =  0  ?  Why  ?  What,  when  m  =  1  ?  Why  ?  What, 
when  p  =  0  ?     Why  ? 

When  there  are  two  similar  radicals,  it  is  best  to  unite  them  in  the 
same  member. 


p    o  ,      ,,  ,.       h-\-hy/\ — 3?        a 

0.  bolve  the  equation, =  — , 

i  +  mv/l  — a;^        » 


An.  x'=+  ^-'{-^'-l')+-^anHI>-m) 


nh  —  am  ' 


^a?  jm^  —  b')  +  2an/j  (6  —  m) 
ni  —  am 


For,  clearing  of  fractions,  we  get  nh  +  nh^l  —  x^  =  ab  +  am 
-v/1  —  x'^,  or,  («i  —  am)  ^/l  —  ic^  =  6  (a  —  n),  or,  (hJ  —  amf 
(l  —  x^)  =  U'(a—ny. 

Developing,  we  get  n~b''  —  2nba7n  +  a-m^  —  x'' (nb  —  ajuf  =  b^a^ 
—lanb^  +  bhi^.    Hence,  a==  {in^—b^)  +  2a?j6  (6— «i)  =  x^{nb  —  amf. 

Hence,      .•  =  +  ^^llMH^IIl^mHS,  ."  =  _ 
{lib  —  ani)  ' 

V a"-  (w^  —  Z.^)  +  2a?i6  (6  —  m) 

7i6  —  am 


7.  Solve  the  equation,  n-^p^  -{■  t?  ■\-  ax=  xy/p^  +  a;*  +  an. 


Ans.  x'  =  +  -v/a'^  — p^  and  a;"  =  —  V  «^  —  P^ 

Verify  these  results.     AYhat  do  they  become  when  a^  =  p2  ?     Why  ? 
What,  when  /  >  a^  ?     Why  ? 

330.  There  is  another  value  which  does  not  appear,  it  is  xz=n. 
For,  by  the  second  principle,  Article  321,  the  rational  terms  in  the  two 
members  must  be  equal.  Equating  them,  we  get  ax  =  an,  or  a;  =  n. 
This  ought  to  be  so,  for  the  given  equation  can  be  put  under  the  form 
of  (n  —  a;)  y/p^  +  x''  —  a  (n  —  x)  =  0,  an  equation  which  can 
plainly  be  satisfied  when  x  =  n.  The  given  equation,  previous  to  the 
24* 


282       EQUATIONS  OF  THE  SECOND  DEGREE. 

division  by  the  common  factor,  n  —  oc,  was  really  a  cubic  equation,  and 
contained  three  values.  We  are  sometimes  enabled  to  detect  a  value,  as 
in  the  above  example,  by  equating  the  rational  factors.  When  the  root 
of  this  equation  will  also  satisfy  the  equation  formed  by  equating  the 
radicals,  it  is,  of  course,  a  value  in  the  given  equation. 


8.  Solve  the  equation,  p-,/x^  —  ct^  +  a  =  n^/ x^  —  a^  +  x. 

(  (P  —  '0'  +  1) 
Ans.  x=^a,  ox  x^=.a  \^ — -. — ~. 

1  —  (i>  —  ») 

331.  There  arc  some  expressions  in  a  fractional  form,  which  must  be 
changed  into  equivalent  fractions  with  rational  denominators. 


n    o  ,      ^1  X-       ni  +  X  +  y/ 2mx  +  x'^ 

9.  Solve  the  equation,  - 


m  +  X —  v^  2mx  +  x^ 

.        ,         (1  +  ^li)'    „         (1  -  JTi)' 

Ans.  X  ^m :=r: ,  X   =  m ^— . 

2(2+  Vw)  2(2— v/n) 

For,  by  multiplying  numerator  and  denominator  of  the  fraction  in 

,     „              ,      ,      T                                    (m -\- X -\- -J  2mx  ■\- x^^"^ 
the  first  member  by  the  numerator,  we  get ^ =^  n. 

Clear  the  equation  of  fractions,  and  extract  the  square  root  of  both 
members.     Then,  m  +  cc  +  s/'^mx  +  a;^r=  d=  m-s/ n,  or  in  (1  dz  Vn) 

—  x=i  —  ^2mx  +  x^.     Squaring  both  members,  we  get  m"^  (1  rt  \/  rif 

—  2mxy\-  ±  V  7()  +  a;'^  =r  2mx  +  x^,  and,  by  transposition,  m^  (1  =h  s/nf 
m(l  ±  ^nf 


2mx  (2  rb  \/n).     Hence,  x  =z 


2(2  ±  Vn) 


10.  Solve  the  e<:|uation,  — ^^^==z z==^^=  =  1. 

^x^  —  a'—^x'^  —  a'' 

Making  the  denominator  rational  as  before,  we  have 


(v/.-+aHV^--a-)-^^^^^^,^^.^,^-^^^,^^,_^,^^^. 


Then,  2^x'  —  a'  =  2a^  —  2x\  or  ^x'  —  a' =  x-  —  al 
Squaring  again,  we  get  x*  —  a'*  =  x*  —  2a;V  +  a*,  or  2a;V  ==  2a\ 
or  x^  =  a^     Hence,  x'  z= -}-  a,  and  x"  =  —  a. 

It  will  be  seen  that  these  values  satisfy  the  equation.  In  this  ex- 
ample, the  second  step  was  not  to  extract  the  square  root  of  both  mem- 
bers, as  in  the  last  example,  because  the  double  product  of  the  radicals 


EQUATIONS  OF  TUE  SECOND  DEGREE.       283 

gave  a  simple  result.  When  the  radicals  are  of  such  a  form  that  their 
double  product  will  not  give  a  simple  result,  it  is  best  to  make  the 
second  step  the  extraction  of  the  square  root  of  both  members. 


11.  Solve  the  equation,  — .  ==-  =  1. 

Ans.  x'  =^  +  h,  x" 


12.  Solve  the  equation,  = — -». 

^x^  +  8m^—    ^x^  —  m^ 

Ans.  x'  =  +  jn,  x"  — 
Sometimes  the  first  step  is  squaring  both  members. 


,001      XT-  X-       n/"  +  ^c  +  v/a  —  X  Im 

13.  Solve  the  equation, = =  \  /  — • 

y/'lm  V     n 

,  '"^        ^a -9       II  ^'l         -: n 

Ans.  X  =^  -\ y/ Ian  —  ??r,  .7:    = y^/'lan  —  m^ 


14.  Solve  the  equation,  ^/a  +  x  +  ^/a  —  x  =  y/la. 

Ans.  x'  =z  +  a,  x"  =  —  a. 


-  _     „  ,        ,                .        v/fl  +  X  +  \/'i  —  X         X 
15.  Solve  the  equation,  - — ■ -^ =  — , 


Ans.  x  =  +  2>/am  —  m^,  x"  =  —  2V am  —  m'. 
What  do  these  values  become  when  m  =  a?     What,  when  w  ^  a  ? 

16.  Solve  the  equation,  ~ =-  = -. 

^  .T  +  1        2x  —  4 

Ans.  x'  =  +  y/b,  x"  =  —  >/5. 

2.7;  4-  4       X 2 

17.  Solve  the  equation,  ~~ p-  =^ , . 

^  '  X  +  2        2x  —  4 

Ans.  x'  =  -\-  2,  x"  =  —  2. 

What  does  the  second  member  become  when  the  first  of  these  values 
is  substituted  ?     How  do  you  explain  the  result. 

18.  Solve  the  equation, " = . 

X  —  o       X  —  a 

Ans.  Both  values  infinite. 
How  are  these  results  explained  ? 


284       EQUATIONS  OF  THE  SECOND  DEGREE. 
nx  +  a         X  +  h 


19.   Solve  the  equation, 


—  b         nx  —  a 


What  do  these  values  become  when  «  =  1  ?     "Why  ?     What,  vyhcn 
7/  >  a\ 

2x  +  1        X  +  2 

20.  Solve  the  equation, = ^- 

*  X  —  2        Zx  —  1 

Ans.  x'=z  +  V^^^,  x"  =  —  \/-^ir 
Why  are  these  results  imaginary  ? 

21.  Solve  the  equation,  ax  -\ =^  —„ a. 

X         x^  —  x 


22.  Solve  the  equation,  2x  + 


Ans. 

2        8a;. 


23.  Solve  the  equation,  ox  -\ = 


Ans.  x'  =  +  2,  x"  =  —  2. 

27ic  — 3       „ 


Ans.  x'  =  +  S,  x"  =  —  3. 


PROPERTIES  OF  INCOMPLETE  EQUATIONS. 

332.  1st.  Every  incomplete  equation  of  the  second  degree  has  two 
values,  and  but  two, and  these  values  are  equal  with  contrary  signs. 

For,  the  general  form  of  the  equation  is  x^  =:zq  or  x^  —  q  =  0.  The 
first  member  may  be  regarded  as  the  difference  of  two  squares,  and  can, 
therefore,  be  placed  under  the  form  of  (x  —  \/ q)  (x  +  n/  q}-  Hence, 
the  equation,  x^  =  q,  may  be  written  (x  —  %/§')  (^c  +  \/q)^^-  Now 
the  product  of  two  factors  being  equal  to  zero,  the  equation  can  be 
satisfied  by  placing  either  factor  equal  to  zero.  Therefore,  x  —  y/q=^Q, 
and  x  -{•  y/q  =  0.  From  which,  we  get  x  =  +  y/q,  and  ar  =  —  -Jq, 
or,  distinguishing  the  values  by  dashes,  a;'  =  +  >/ q,  x"  z=^  —  ^q. 
Since  there  are  but  two  factors,  the  equation  can  be  satisfied  in  but  two 
ways.  Hence  there  are  but  two  values,  and  Ave  sec  that  these  are  equal 
with  contrary  signs. 

By  solving  directly  the  equation,  x^  =  q,  we  would  obtain  the  same 
results.     But  the  process  we  have  adopted  is  to  be  used  hereafter  for 


EQUATIONS    OF    THE    SECOND    DEGREE.  285 

complete  equations,  and  it  is  well  to  know  that  the  properties  of  com- 
plete equations  are  also  those  of  incomplete  equations. 

2d.  Every  incomplete  equation  of  the  second  degree  can  be  decom- 
posed into  two  binomial  factors  of  the  first  degree  with  respect  to  .r ; 
the  first  factor  being  the  algebraic  sum  of  x  and  the  first  value  with  its 
sign  changed,  and  the  second  factor  being  the  arithmetical  sum  of  a- 
and  the  second  value  with  its  sign  changed. 

The  factors  have  already  been  obtained,  and  are  (x —  V  q),  an  J 
(x  +  %/(/);  and  we  see  that  these  correspond  to  the  enunciation  of  the 
second  property.  The  product  of  these  factors  is  zero.  Hence,  when 
wc  know  the  two  values  of  an  incomplete  equation,  we  can  always  form 
the  incomplete  equation  itself  which  gave  those  values.  We  have  only 
to  change  the  signs  of  the  values,  and  connect  them  with  x.  We  will 
then  have  the  binomial  factors,  and,  placing  their  product  equal  to  zero, 
we  will  have  the  equation  required.  Thus,  form  the  equation  that 
gives  for  x  the  two  values,  -f  2  and  —  2.  The  two  factors  are  (x  —  2) 
and  (x  +  2),  and  the  equation  is  (x  —  2)  (a;  -f  2)  =  0,  or  x^  —  4  =  0. 
The  result  can  be  verified  by  solving  the  equation,  x'^  —  4  =  0.  Or, 
since  in  the  equation,  (x  —  2)  (x  +  2)  =  0,  we  have  the  product  of  two 
factors  equal  to  zero,  the  equation  can  be  satisfied  by  placing  either 
factor  equal  to  zero.  Hence,  x  —  2  =  0  and  .r  +  2  =:  0.  And  these 
equations,  when  solved,  give  the  values,  +  2  and  —  2. 


EXAMPLES. 

1.  Find  the  equation  that  gives  for  x  the  two  values,  -f  a^,  and  —  a^. 

Ans.   x^  —  a'  =  0. 

2.  Find  the  equation  whose  values  arc    +  -/oi,  and  —  y/ab. 

Ans.  x^  —  ai  =  0. 

3.  Find  the  equation  whose  values  are  -f  o  —  I,  and  -f  h  —  a. 

Ans.  x'  +  2ah  —  1/  —  a^=0. 


4.  Form  the  equation  whose  values  are  -f  y/ m  —  a,  and  —  ^  m  —  a. 

Ans.  jp-  -f  a  —  m  =  0. 


5.  Form  the  equation  whose  values  are  +  ^f  m  —  a —  s/ in^  —  a^ 
and  —  \/  in  —  a -{-  ^m^  —  a^. 

Ans.  x^  —  (m  —  a)  —  (m^  —  a^)  -f  2^w— a  ^vv'  —  a'=  0. 
Verify  this  result  by  solving  the  equation. 


286       EQUATIONS  OP  THE  SECOND  DEGREE. 


6.  SolvG  tho  equation  whose  values  are  a  —  ^Z  d^  —  m?,  and  —  a 
+  ^a?  —  mK  Ans.  x^  —  2a?+m?+2a  v/m^  —  a"  z=  0 , 

Verify  this  result  by  solving  the  equation. 

7.    Form    the    equation    whose    values    are    -f-   —  ,  and 

_  ^iiv"  +  a" 


8.  Form  the  equation  whose  values  are  —  ■ ,  and 


Ans.  (m^  —  ?i^)  cc'^  =  an^. 

9.  Form  the  equation  whose  values  are  w?  —  n^,  and  —  rti^  -j-  n^. 

Ans.  x^  =  m*  —  2m^n^  +  %*. 
What  do  the  values  become  when  m  =  ?i  ?     "What,  when  m  =  0  ? 

BINOMIAL  EQUATIONS. 

333.  Any  binomial  equation  of  the  n^^  degree  can  be  solved  as  the 
preceding  equations.  We  have  only  to  extract  the  h*""  root  of  both 
members. 

Let  a;^  =  27;  then,  x  =\/W  =  3.     Let  x""  =  q;  then,  x  ==  ^~q. 

If  n  be  an  even  number,  the  radical  must  have  the  double  sign,  and 
two  values,  at  least,  of  the  equation  will  be  known.  If  n  be  an  odd 
number,  only  one  value  will  be  known.  It  will  be  demonstrated  here- 
after, that  every  equation  of  the  ?i*''  degree  has  n  values.  The  method 
of  finding  the  other  values  will  then  be  explained. 


Let  a;"  =  a^ ;  then,  x  =  ±yd^  =  ±\/  ^/^(^  =  ±n/±  a. 

Then,  x' =  +  ^  +  a,  x"  =  —^-\-a,  x'"  ^  +  ^"-^a,  xJ'"  =  — 
v/— a. 

In  this  example  the  four  values  of  x  have  been  determined,  but  it 
is  not  often  the  case  that  they  all  can  be  found. 


Let  x^  =  «%•  then,  x  =  ±  V*'  =  zfc^Y  v^'t?  =  ±  V±  a,  and 
four  of  the  six  values  are  known. 


EQUATIONS    OF    THE    SECOND    DEGREE.  287 


Let  a;'2  =  a%-  then,  x  =  ±'y^  =  ±\J l/cc"  =  ±i/a,  and  only 
two  of  the  twelve  values  are  known. 


EXAMPLES. 

1.  Solve  the  equation,  ^x*  —  6az  +  oa^  =  z  —  2a. 

Ans.  x'  —  +  ^z  +  a,  x"  =  —  ^z  '\-  a,  x'"  =  +  ^117(7+^, 
x"" v/—  (z  +  a). 

2.  Solve  the  equation,  x*  =  256.         Ans.  x'  =  +  -i,  x"  =  —  4. 

3.  Solve  the  equation,  ^a^  +  x^  +  y/a}  —  x^  =  ni^2. 

Ans.  a/  =  +  i/a*  —  {m?  —  a^f,  x"  =  —i/a*  —  (m''—a^f. 

4.  Solve  the  equation,  a;'"  =  32. 

Ans.  x'  =  +  ^%  x"  =  —  ^2^ 

5.  Solve  the  equation,  -^^-^ _:  ^ =2^^-- 

s/m 

Ans.  x'  =  +  j9  Vm  (2  —  m),  x"  =  — j^l/m  (2  —  m). 

Qj  Solve  the  equation,  x*  =  1. 

Ans.  x'=-\-l,  x"  =  —  l,  x'"  ±  +  v/^=T,  ./■""  =  — vZ-'^nr 

Verify  the   preceding   results  by  substituting   them  in  the  given 
equations. 

GENERAL  PROBLEMS  IN  BINOMIAL  EQUATIONS. 

334.  1.  The  sum  of  two  numbers  is  «,  and  the  ratio  of  their  squares 
m  :  what  are  the  numbers  ? 

Ans.   X  = r=r,  X  = :^ 

•  1  +  v/m  1  —  -v/"^ 

For,  let  X  =  one  number ;  then,  a  —  x  =  the  other,  and,  by  the 

x^ 
conditions, -,  =  in.    Now,  extract  the  square  root  of  both  mem- 

(a  —  xy  '  ^ 

x' 


ziz  y/m.     Hence,  • ,  r=  -\-^m,  and 

11       ^  a  —  X, 

—  ^/n.      From    the  first  equation,  u^  ^=  ■\-  a-^m  —  x^  y/m, 


a  —  X 
or 


288       EQUATIONS  OF  THE  SECOND  DEGREE. 

—     ,         +  a  »/«i.    ,,      ,        _ 

x'(l -^  .y^n)  =a  yin,  or  x' zr= r=.      l^rom  the  second  equa- 

1  H    ^m 

tion,  we  get  x"  r=  —  a  y/  m  +  x"^m,  or  x"  =i z;^.      Now,   if 

1  —  y/m 
m  =  \,  tlic  second  value  is  infinite.     How  is  this  result  to  be  inter- 
preted ?     By  going  back  to  the  equation  of  the  problem,  we  see  that 

x^ 
it  becomes,  under  this  hypothesis, -^  =  1 ;  an  equation  which 

can  only  be  true  when  a  z=.  0,  or  when  x  ^=  —  .     The  first  supposition 

contradicts  the  enunciation,  the  second  is  the  value  found  for  x'.  The 
second  value  can  only  exist  when  there  is  a  contradiction  to  the  state- 
ment :  it  ought,  then,  to  appear  under  the  symbol  of  absurdity.  We 
may  explain  the  result  otherwise;  thus:  when  m  =  1,  the  equation  in 

x"  becomes  x"  —  x"  =z  —  a,  or  • j, =  +  —^,   or  0  =  -| ^,. 

This  equation  is  plainly  absurd  for  any  finite  value  of  x". 

The  value,  x"  =  —  --,  satisfies  the  equation  of  the  problem,  as  it 


manifestly  ought  to  do.     We  have —  =  1,  or  — —  =  1, 

IF 


(a\         '  "'  (Oa  —  aY 


or,  -^[  =  ^^^^^,  or  O'a'  =  0'  (Oa  —  af,  or  0  =  0. 

2.  The  sum  of  two  numbers  is  a,  and  the  ratio  of  their  fourth  powers 
is  m.     What  are  the  numbers  ? 

,  a  ij  m        „  a  yj  m 

1  +  V?u'  1  —  Xjm 

3.  Two  numbers  are  to  each  other  as  m  to  «,  m  being  greater  than  n, 
and  the  ratio  of  their  squares  is  equal  to  (.^  divided  by  the  square  of  the 
greater.     What  are  the  numbers  ? 

^.  an       „         ,      ,     ai^ 

Ans.  First,  ± ,     Second.  ±  — 5-. 

m  '         vv 

4.  Two  numbers  are  to  each  others  as  m  to  n,  and  the  sum  of  their 
fourth  powers  is  equal  to  a".     What  are  the  numbers  ? 


Ans.  First,  ± :,     Second,  ±  -- 


EQUATIONS    OF    THE    SECOND    DEGREE.  289 

5.  Two  numbers  are  to  each  other  as  ni  to  n,  and  the  diflforeuco  of 
their  fourth  powers  is  O*.     What  are  the  numbers  ? 

A71S.  iirst,  it  ,  Second, 


y  m* — n*  %/  «t*  — ,!•• 

What  do  these  values  become  when  m  ~  n  ?     Why  ?     What  when 
m=0?     Why? 

6.  The  cube  root  of  twice  the  square  of  a  number  is  2.      What  is 
the  number  ? 

Ans.  Either  -|-  2,  or  —  2. 

7.  The  cube  root  of  a  times  the  square  of  a  number  is  h.     Wliat  is 
the  number? 


Ans.  X 


'-Wl--\/l 


What  do  these  values  become  when  a  =  0  ?     What  when  h  =0? 

8.  The  square  of  a  number  multiplied  by  the  first  power  of  the  same 
number  is  equal  to  64.     What  is  the  number  ?  Ans.  4. 

9.  A  man  put  out  a  certain  sum  of  money  at  G  per  cent,  interest. 
The  product  of  the  interest  upon  the  money  for  G  months  by  the  interest 
for  4  months  was  600  dollars.     What  was  the  sum  at  interest  ?  , 

A71S.  81000. 

10.  A  man  put  out  a  certain  sum  of  money  at  6  per  cent,  interest. 
The  product  of  the  interests  upon  it  for  3,  6  and  9  months  was  820  i. 
What  was  the  sum  ?  Ans.  6100. 

11.  The  successive  quotients,  of  a  quantity  divided  first  by  a,  and 
then  by  h,  will,  when  multiplied  together,  give  a  product  equal  to  mn. 
What  is  the  quantity  ? 

Ans.  a;'  =  +  ^ahmn,  x"  =  —  y/nhmn. 

COMPLETE   EQUATIONS   OF  THE    SECOND   DEGREE. 

335.  The  most  general  form  of  such  eciuations  is, 1 —  n=-r; 

b        m 

clearing  of  fractions,  we  get,  —  amx^  +  hex  —  Imn  =  hmr.     Dividing 

by  the  coefficient  of  x^, —  am,  we  have  x'  — 1 = :    and 

am        a  a 

by  transposition,  x^ == ^ \     Now,  the  second  member 

am  a 

25  T 


290  EQUATIONS    or     THE    SECOND    DEGREE. 

may  be  plainly  represented  by  a  single  letter,  —  q,  and  tlie  coefficient  of 
the  second  may  be  represented  by — p.  Hence,  the  given  equation  as- 
sumes the  form  of  x^  — px  =  —  q;  in  which  the  highest  term  of  the 
unknown  quantity  has  a  coefficient  plus  unity.      Had  the  coefficient, 

—  -  of  a;^  in  the  original  equation  been  affected  with  the  positive  sign,  it 

would  have  reduced  to  the  form  of  x^  +  2^-^  —  5-  ^^^  ^^^  coefficient 
of  x^  been  positive,  and  that  of  x  negative,  the  equation  would  have 
assumed  the  form  of  x^ — px  ^  q.  Had  the  last  conditions  been  ful- 
filled, and  the  prevailing  sign  in  the  second  member  at  the  same  time 
negative,  the  equation  would  have  assumed  the  form  of  x^ — px  =  —  q. 
And,  since  every  change  that  may  be  made  upon  the  signs  of  the 
coefficients  of  x^  and  x,  and  upon  the  signs  of  the  known  terms,  will 
eventually  lead  us  to  one  of  the  preceding  forms,  we  conclude  that 
every  complete  equation  of  the  second  degree  may  be  placed  under  one 
of  the  following  forms  : 

x^  +  px  =  q,      First  form, 
a;2  — jyx  ^  -\-q,  Second  form, 
x^  +  pjx  =  —  q,  Third  form, 
x^  — px  ^=  —  q,  Fourth  form. 

When  a  complete  equation  is  given  to  be  solved,  it  must  first  be  placed 
under  one  of  these  forms.  To  do  this,  clear  it  of  fractions,  if  it  con- 
tain any,  and  then  make  the  coefficient  of  the  first  term  plus  unity,  if 
not  already  so,  by  dividing  by  this  coefficient.  The  resulting  equation 
will  then  be  under  one  of  the  required  forms.  The  form  that  the 
resulting  equation  assumes  will,  of  course,  depend  upon  the  prevailing 
signs  in  the  given  equation. 

x^       Ix 
Reduce  the  equation, —  ^  8  =  ^ — ■  -4j  to  one  of  the  four  forms. 

Clearing  of  fractions,  we  have  cc^ — 2a;+12=— 6,  or  .r^ — 2x= — 18. 
Hence,  the  equation  has  assumed  the  fourth  form. 

Reduce  the  equation, —  +  o  =  —  f  to  an  equivalent  equation, 

which  will  appear  under  one  of  the  four  forms. 

Multiplying  both  members  by  20,  the  least  common  multiple  of  the 
denominators,  we  will  get  5x^  —  lOx  -f  00  =  —  24.  Now,  divide  by 
-|-  5,  to  make  the  coefficient  of  x^  plus  unity.     Then,  x^  —  2x  +  12 

24 

^__or-4f 


EQUATIONS    OB'    THE    SECOND    DEGREE.  291 

We  see  in  this  particular  case,  that  we  might  have  obtained  the  sam-e  re- 
sult by  multiplying  the  original  equation  by  4.  This -would  have  made  the 
coefficient  of  the  first  term  unity,  which  is  the  main  point  to  be  attended 
to ;  then,  by  transposition,  the  equation  would  have  become  x^  —  2x  = 
— 16|.  And  we  see  tha-t  the  equivalent  equation  is  of  the  fourth  form. 
It  frequently  happens  that  it  is  impossible  to  make  all  the  terms  entire, 
and  the  coefficient  of  the  first  term  at  the  same  time  plus  unity.  But, 
since  this  coefficient  must  always  be  plus  unity,  we  derive  the  following 
rale  for  reducing  any  equation  to  one  of  the  proposed  form. 


RULE. 

Multiplij  both  members  of  the  equation  by  the  reciprocal  of  the  co- 
efficient of  the  first  term,  and  then  transjxjse  all  the  known  terms  to 
the  second  member. 

The  multiplier  in  the  equation,  |  o;^  —  2.r  =:  4  is,  +  |.  The  multi- 
plier in  the  equation,  —  |  cc^  —  2x  =  4  is,  —  -^ . 

The  reason  of  the  rale  is  obvious,  and  needs  no  explanation. 


EXAMPLES. 

1.  Keduce  the  equation,  |  x^  —  J  a;  +  G  =  4,  to  one  of  the  four  forms. 

Ans.  X?  —  -=  —  3. 

2.  Keduce  the  equation,  —  |  a;^  —  \x  +  6  =  -j-  4,  to  one  of  the 

four  forms.  ,  x 

Ans.  x^+-  =  +  ^. 

3.  Reduce  the  equation,  ^'^  x"^  +  lx=zl,  to  one  of  the  four  forms. 

Ans.  x^  -\-  ^x  =  25. 

5 

Ans.  x^  —  bx  =  25. 

5.  Reduce  the  equation,  —  f  .r'^  —  J  j'  -f  fi  =  +  4,  to  one  of  the  four 

forms.  X 

Ans.  x^  +  '-  =  o. 

336.  If  the  first  member  of  the  equation  put  under  one  of  the  four 
forms  is  a  perfect  square,  the  solution  can  be  as  readily  effected  as  in 
the  case  of  an  incomplete  equation.     For,  we  will  only  have  to  extract 


292  EQUATIONS    OF    THE    SECOND    DEGREE. 

tlic  square  root  of  botli  mciubcrs,  and  then  transfer  the  known  term  or 
terms  in  tlie  first  member  to  the  second  member,  and  then  the  solution 
will  be  complete.  Suppose  that  the  equation  is  a;^  +  2aa;  +  a^  =&. 
Extract  the  root  of  both  members,  then  a;  +  a  =  =b  ^b.  Hence, 
x'  =  —  a  -\-  -s/h,  and  a;"  =  —  a  —  -Jh.  We  see  that  the  equation 
has  two  values,  and  that  these  are  not  numerically  equal,  as  in  the  case 
of  incomplete  equations. 

Now  if,  by  any  artifice,  we  can  make  the  first  member  a  complete 
.square,  it  is  plain  that  there  will  be  no  difficulty  in  the  solution  of  a 
compl-ete  equation  of  the  second  degree.  Let  us  then  assume  the  equa- 
tion, x^  -\-  "px  =  q,  and  examine  what  modification  the  equation  must 
undergo,  in  order  that  its  first  member  may  be  made  a  perfect  square. 
The  square  of  a  binomial  is  composed  of  the  square  of  the  first  term, 
plus  the  double  product  of  the  first  by  the  second,  plus  the  square  of 
the  second  term.  The  square  of  a  binomial  is,  therefore,  a  trinomial. 
.ir^  +  jjx  must  then  be  made  a  trinomial  by  the  addition  of  some  quan- 
tity before  the  first  member  \Vill  become  a  perfect  square.  What  is  the 
quantity  to  be  added  ?  Take  the  expression,  (x  +  a)^  =  x^-\-  2ax  -f-  a'. 
We  see  that  the  third  term  of  the  trinomial  is  the  square  of  the  second 
term  of  the  binomial  in  the  first  member ;  and  we  sec  that  2a,r,  the 
second  term  of  the  trinomial,  when  divided  by  2x,  twice  the  first  term 
of  the  binomial,  gives  a  quotient,  a,  which  is  the  second  term  of  the 
binomial.  This  quotient,  squared,  is  the  third  term  of  the  trinomial. 
Now,  suppose  we  only  knew  the  first  two  terms  of  the  trinomial,  x^  and 
2ax,  and  wished  to  ascertain  what  was  the  binomial,  which,  when 
squared,  would  give  these  for  the  first  two  terms  of  its  square.  We 
would  know  that  the  first  term  of  the  binomial  must  be  x  ;  and,  since  2ax, 
the  second  term  of  the  trinomial  is  twice  this  first  term  by  the  second 
terra,  it  is  plain  that  the  second  term  can  be  found  by  dividing  by  2a,',  twice 
the  first  term.  Having  found  a,  the  second  term  of  the  binomial,  we 
have  only  to  square  it,  and  the  third  term  of  the  trinomial  will  be 
known. 

Apply  these  principles  to  the  expression,  a;^  -f  px,  regarded  as  the 
first  two  terms  of  a  trinomial  that  is  a  complete  square.  It  is  plain 
that  X  is  the  first  term  of  the  binomial,  the  first  two  terms  of  whose 
square  are  x^  +  ^«.     The  second  term  of  the  binomial  can  be  found 

by  dividing  j^x  by  2x,  twice  the  first  term  :  the  quotient,  ~,  is  the 
second  term  of  the  binomial,  and  its  square,  ^,  is  the  required  tliird 


EQUATIONS  OF  THE  SECOND  DEGREE.       293 

term  of  the  trimonial.      If  then  ^  be  added  to  x^  -\-  V^,  the  first 

member  will  be  a  perfect  square.  But,  if  we  add  ^  to  the  first  mem- 
ber, we  must  also  add  it  to  the  second,  eke  the  equality  of  the  two 
members  will  be  destroyed.     Hence,  the  equation,  x^  -\-  px  =^  q,  can 

be   changed  into  the  equivalent    equation,  x-  +  px  +  x  ~  2  "^  jT" 
in  which  the  first  member  is  a  perfect  square. 
Now,  extract  the  square  root  of  both  members, 

X  +  ^  =  ±  \/y  +  q.      Hence,  x'  =  —  ^  +  ^/^  +  q,  and 

Either  of  these  values  when  substituted  for  x  will  satisfy  the  equation. 
The   first,    when    substituted,    will   give    ( —  ^"'"V/'T'^!?^"^ 

^  ("~f  +  \/f  +  '^)  =  *?'  ""'  ^-^  \/f'  +1+J  +  1- 


'h^'sji 


+  <i 


Reducing  the  first  member  we  have,  q  =  q.  Hence,  the  first  value 
satisfies  the  equation.     The  second,  when  substituted,  gives 

/f  p^  p'         //>' 

Py  t  +  2  +  T  "^  '^  ~  2  ~^'V  J  +  'i  =  !?>  or  !?  =  q. 

Hence,  the  second  value  also  satisfies  the  equation. 

We  have  taken  an  equation  under  the  first  form,  but  it  is  obvious 
that  the  principles  deduced  are  applicable  to  equations  under  either  of 
the  three  other  forms.  Hence,  for  the  solution  of  a  complete  equation 
of  the  second  degree,  we  have  this  general 

RULE. 

Reduce  the  equation  to  one  of  the  four  forms,  hy  multiplying  hath 
members  hy  the  reciprocal  of  the  coefficient  of  the  highest  power  of  the 
unknown  quantity,  and  by  transposing  all  the  knoicn  terms  to  the  second 
25* 


291       EQUATIONS  OF  THE  SECOND  DEGREE. 

■memher.  Next,  add  the  square  of  half  the  coefficient  of  tlie  fir&t  poiccr 
of  the  unknmon  quantity  to  loth  memhers,  and  then  extract  the  square 
root  of  hoth  memhers. 

Finallij,  separate  the  unhnown  quantity  from  the  knoion  quantities 
hy  leaving  it  alone  in  the  first  memher. 

A  common  error  with  beginners  is,  to  complete  the  square  without 
observing  whether  the  coefl&cient  of  the  first  term  is  plus  unity.  But 
it  will  be  seen  that  the  rule  requires  the  first  step  to  be  the  making  of 
this  coefiacient  plus  unity,  if  not  already  so. 


EXAMPLES. 

1.  Solve  the  equation,  —  x^  +  16a;  =  28. 

Ans.  x'  =  +  2,  x"  =  +  14, 

2.  Solve  the  equation,  x^  —  4:X  =  — 4. 

Ans.  x'  =  +  2,  x"  =  +  2. 

3.  Solve  the  equation,  x^  +  4a;  =  —  4. 

Ans.  x'  =  —  2,  x"  =  —  2. 

It  is  generally  best  to  reduce  the  terms  under  the  radical  to  a  common 
denominator. 

Take  the  equation,  —  ax^  —  cx=  —  m.     This,  when  solved,  gives 

X  =  —  -— d=\/ 1-  T— 0-     The  two  terms  under  the  radical  can  be 

2«        V     a        4a'' 

reduced  to  a  common  denominator  by  multiplying  the  numerator  and 
denominator  of  the  first  by  4a,  If  m  had  been  divisible  by  a,  then  a 
would  not  appear  in  the  denominator  of  the  first  term,  and  the  multi- 
plier then  would  have  been  4a^.  The  single  term,  into  which  all  the 
known  quantities  in  the  second  member  have  been  collected  after  the 
(Coefficient  of  x^  has  been  made  plus  unity,  is  called  the  ahsolute  term. 

In  the  preceding  equation,  —  is  the  absolute  term.    Then,  to  reduce  the 

terms  under  the  radical  to  the  same  denominator,  we  multiply  numerator 
and  denominator  of  the  absolute  term  by  either  4  tim.es  the  coefficient 
of  the  second  power  of  x,  or  4  times  the  square  of  this  coefficient. 
This  rule  is  only  applicable  when  the  reduced  term  in  the  second 
member  was  entire,  previous  to  making  the  coefficient  of  x^  plus  unity. 


EQUATIONS  OF  THE  SECOND  DEGREE.       295 

4.  Solve  the  equation,  2x^  —  Sx  =  5. 

Ans.  x'  =  +  2J-,  x"  =  —  1. 


For,  solving,  we  get,  x  =  |  ±^^  +  J>^  =  ^dtz  v/|g  +  /g  =  |  ± 
1=  -\-2h,ov-l. 

5.  Solve  the  equation,  2x^  —  Sx  =  2. 

Ans.  x'  =  +  2,  x"  =  —  A- 

For,  solving,  we  get,  x  =  |  ±  N/T+7g  =  |  ±  v/  -^"^  =  |± 
I  =  +2,  or— *. 

,.        CI     1  -1  ,•  -^^  ^'^  "^  ^'* 

o.  bolve  the  equation,  — ,  H =  -—  +  — . 

wr        c        ?>!''        c 

^i??s.  x  =  +  o,  X    =  —  a . 

c 

337.  These  few  examples  are  given  to  show  how  complete  equations 
can  be  solved.  But  the  solutions  can  be  better  understood  when  some 
of  the  properties  of  these  equations  arc  known. 


First  Projjcrf^. 

Every  complete  equation  of  the  second  degree  has  two  values,  and 
but  two,  for  the  unknown  quantity. 

For,  resume  the  equation,  x^  +  px  =  q.     Complete  the  square  of 

the  first  member,  and  we  have  x'  -\- px  +  -^  =  q  +  ^.     Now,  the 

4  4 

first  member  being  a  square,  the  second  member  may  be  represented  by 

a  square  also.     Hence,  x'^  -}-  j^x  -{-—-=  ni^,  or  (x  +  j^Y  —  "^^  =  ^• 

Now,  the  first  member  being  the  diflerenee  of  two  squares,  may  be  re- 
solved by  the  principle  of  the  sum  and  difi"erence  into  two  factors. 
Then  the  equation,  (x  +  pY  —  m^  =  0,  will  be  changed  into  the  equiva- 
lent equation,  (x  -i-  p -{- m)  (x  -{- p  —  m)  =  0.  And  since,  when  the 
product  of  two  factors  is  equal  to  zero,  the  equation  can  be  satisfied  by 
placing  either  factor  equal  to  zero,  we  have,  x  +  p  -\-  m  =  0,  and 
x-\-p  — m  =  0.  From  which  we  get  X  =  — p  —  m,  iindx  =  — p  +  7n  ; 
or,  distinguishing  the  values  by  dashes,  x'  =  — p  —  m,  and  x"  =z  — 
p  +  m.     Now,  if  we  replaced  m  by  its  value,  we  would  have  the  solu- 


-UC)  EQUATIONS    OF    THE    SECONIJ    DEGREE. 

tions  previously  obtained.  We  sec  that  there  are  two  values  nume- 
rically unequal,  and,  since  the  equation,  (x  +  p -\-  m)  (cc  -f  7)  —  m)  =  0, 
contains  but  two  factors,  it  can  be  satisfied  in  two  ways,  and  only  in  two 
ways.  Hence,  there  are  two,  and  only  two,  values  for  the  unknown 
quantity. 

Sfcond  Pmpcrty. 

338.  The  first  member  of  every  equation  of  the  second  degree  can  be 
decomposed  into  two  factors  of  the  first  degree  with  respect  to  x^  the 
first  factor  being  the  algebraic  sum  of  x  and  the  first  value  of  x  with 
its  sign  changed )  and  the  second  factor  being  the  algebraic  sum  of  x 
and  the  second  value  with  its  sign  changed.  The  second  member  of 
the  equation  after  this  decomposition  will  be  zero. 

The  factors  have  already  been  obtained,  and  are  (a^  +  p  -f  7)1)  and 
[x,  -\r  p  —  ■?«)•  By  comparing  these  factors  with  x'  and  x" ,  we  see 
that  Ar  p  -\-  in  is  equal  to  —  x' ,  and  that  +  p  —  «i  is  equal  to  —  a;", 
and  these  factors  then  can  be  changed  into  the  equivalent,  {x  —  cp') 
and  {x  —  a;")  ;  and  the  equation,  {x  ■\-  p  -\-  m)  (x.  +  j)  —  m)  =  0,  can 
be  changed  into  (x  ■ —  x')  {x  —  x")  =  0. 

339.  This  important  property  enables  us  to  find  the  equation  whose 
values  are  known.  We  have  only  to  change  the  sign  of  the  first  value, 
and  prefix  -f  a?  to  it,  and  the  first  factor  will  be  known;  and  then  to 
change  the  sign  of  the  second  value,  and  prefix  +  x  to  it,  and  the 
second  factor  will  be  known.  The  product  of  these  factors  placed 
equal  to  zero  is  the  equation  required. 

Find  the  equation,  whose  values  are  -f  2  and  —  3.  The  first  factor 
must  be  (x  —  2)  ;  the  second  factor  must  be  a;  +  3.  Hence,  (x  —  2) 
(x+3)  =  0,  is  the  equation  required.  Expanding  the  first  member,  we 
have  x^  -{-  X  —  6  =  0,  or  a;^  -f  a;  =  6.  Completing  the  square,  and 
solving,  we  get  a-  =  —  J  db  ^Q  +  I  =  —  i  rt  v/-^'  =  —  J  d=  |  = 
-f  I,  or  +  2,  or  —  |  =  —  3.  Hence,  the  process  is  verified.  But 
the  verification  might  have  been  made  more  readily,  thus :  the  equation 
(x  —  2)  (.r  4-  3)  =  0,  can  be  satisfied  by  placing  either  factor  equal  to 
zero.  Hence,  x  —  2  =  0,  and  a;  +  3  =  0.  These  equations  give  the 
preceding  values,  +  2  and  —  3. 

A  few  examples  will  make  the  beginner  more  familiar  with  the  second 
property. 


EQUATIONS    OF    THE     SECOND    DEGREE.  291 


EXAMPLES. 


1.  Form  the  equation  whose  values  are  both  —  2. 

Ans.  (x  +  2)  Qc  +  2)  =r  0,  or  a-^  4.  4.^  a.  4  =  0. 

2.  Form  the  equation  whose  values  are  (f  +  ^h,  and  a  —  ^/i. 
Ans.  (x  —  a  —  ^h)  (x  —  a+  ^V)  =  0,  or  x^  —  2«a-  =  h  —  a^. 


3.  Form  the   equation  whose  values  are  — 2m  +  y/n  +  Ani^,  and 
-  2w?  —  y/7i  +  4ml  Ans.  x^  +  Amx  =  n. 


4.  Form  the  equation  whose  values  are  a  —  n+  s/m  —  2an  +  n'^, 
and  a  —  n  —  ^m  —  2an  +  ?i^. 

Ans.  x^  —  2  (a  —  «)  ./;  =  m  —  a^. 
Verify  these  results  by  solving  the  equations. 

340.  The  second  property  of  complete  equations  sometimes  enables  us 
to  solve  an  equation  very  readily.  For,  whenever  the  factors  of  an  equa- 
tion are  apparent,  we  need  not  solve  the  eciuation  itself,  but  only  those 
factors  separately  placed  equal  to  zero.  Thus,  take  the  equation 
01?  +  Ix  =  ax.  By  transposition  we  get,  x^  ■\-  hx  —  ax  =  0.  We 
see  that  a;  is  a  common  factor  to  the  first  member,  and  the  equation 
may  be  written  x  (x-\-  h  —  a)  =  0.  And  since,  when  the  product  of 
two  factors  is  equal  to  zero,  the  equation  can  be  satisfied  by  placing 
either  equal  to  zero,  we  have  a;  =  0,  and  x-\-  h  —  a  =  0.  Hence,  the 
two  values  are  0,  and  a  —  h.  The  general  equation,  x"^  -{■  hx  —  ax 
=z  0,  slioios,  furthermore,  that  when  the  unknown  quantity  enters  into  all 
the  terms  of  the  first  member  of  an  equation,  ichose  second  member  is 
zero,  one  value  of  the  iinknown  quantit}/  must  nlwai/s  be  zero. 

Take,  as  a  further  illustration  of  the  use  of  the  second  property,  the 
equation,  ax^  —  bx^  -[-ax  —  bx  =  bx  —  ax.  This  equation  can  be 
written  (a  —  b)  x^  +2  (a  —  b)  x  =  0,  ot  x  {  (a  —  b) x  +  2  (a  —b)  ) 
=  0,  or  X  (a  —  b)  (x  +  2)  =  0,  or  x(a  —  b)  =  0,  and  x  +  2  =  0. 
Hence,  x'  =  0,  and  x"  =  —  2. 

Again,  x^  —  ax  -\-  x  —  a  =  0,  may  be  written  x  (x  —  a)  -f  ■'"  —  ('' 
=  0,  or  (x  —  rt)  (x  -f-  1)  =  0.     Hence,  x'  =  a,  and  x"  =  —  1. 

We  have  seen  that  the  first  two  properties  of  complete  equations  are 
also  properties  of  incomplete  equations.  The  remaining  properties  are 
also,  as  we  shall  see,  common  to  both  classes  of  equations. 


298  EQUATIONS    OF    THE     SECOND    DEGREE. 


Tliird  Properti/. 

341.  The  algebraic  sum  of  the  values  of  every  complete  equation  of 
the  second  degree  is  equal  to  the  coefficient  of  the  first  power  of  the 
unknown  quantity,  with  its  sign  changed. 

For  the  equation,  si:^  +  px  =  q,  when  solved,  gave  the  two  values, 


^'  =  -f  +  \/^ 


+  <1 


^'  =  -f-\/^+- 


Adding  these  equations,  member  by  member,  we  get  x'  +  x"  —  — p, 
as  enunciated. 

The  third  property  of  complete  equations  is  also  a  property  of  incom- 
plete equations.  For,  an  incomplete  equation  of  the  form,  x^  =  q,  may 
be  written,  x^  ~\-0x  =  q.  The  two  values  of  this  equation,  solved 
either  as  a  complete  or  as  an  incomplete  equation,  are,  cc'  =  -f  ^q^  and 
x"  =  —  ^q.  The  sum  of  these  values,  x'  +  x",  is  plainly  zero, 
which  is  the  coefficient  of  the  first  power  of  x.  An  incomplete  equa- 
tion then  may  be  regarded  as  a  complete  equation,  whose  two  values 
are  numerically  equal,  but  affected  with  contrary  signs. 


Fourth  Pfop)erty. 

342.  The  product  of  the  values  in  every  complete  equation  of  the 
form,  x^  +  P^  =  (J^  is  equal  to  the  second  member  or  absolute  term 
with  its  sign  changed. 


For  these  values  are  cc'  =  —  "o"  +  \/  T"^  S' 

andx"  =  —^  —  S^/^+q. 

and  their  product,  x'x"  =  -{-  ^ U-  -\-  q\  = 


I- 


This  property  belongs  also  to  incomplete  equations,  for  the  two  values 
of  the  equation,  x^  =  q,  are  x'  =  +  y/q  and  x"  =  —  .^q,  and  their 
product,  x'x"  =■  —  q. 


EQUATIONS    OF    THE     SECOND    DEGREE.  299 

"We  will  see,  hereafter,  that  the  third  and  fourth  properties  belong  to 
equations  of  every  degi-ee,  and  that  the  first  and  second,  with  some 
modifications,  are  also  properties  of  all  equations. 


EXAMPLES. 

1.  What  is  the  product,  and  what  the  sum,  of  the  values  in  the 
equation,  x^  +  4x  =  — 4.  Aiis.  x'x"  =  +  4,  x'  +  x"  =  — 4. 

2.  What  the  product,  and  sum,  in  the  equation,  x^  +  x  =  1. 

Ans.  x'x"  =  —  1,  x'  +  x"  =  —  1. 

3.  What  the  product,  and  sum,  in  the  equation,  x^ — px  =  —  tj. 

Ans.  x'x"  =  -\-  q,  x'  +  x"  =  +  p. 


Fifth  Property. 

343.  The  value  of  x,  in  every  complete  equation  of  the  second 
degree,  is  half  the  coeflScient  of  the  first  power  of  x,  plus  or  minus  the 
square  root  of  the  square  of  half  this  coefficient  increased  by  the 
absolute  term. 

For  the  equation,  x^  +  px  =  tp  when  solved,  gives  x  =  —  ^  ± 
\  /  -J  +  q^  which  agrees  with  the  statement. 

It  is  obvious  that  this  is  also  a  property  of  incomplete  equations, 
since,  for  such  equations,  ;?  =  0. 

We  have  shown  the  foregoing  properties  by  operating  upon  an  equa- 
tion of  the  first  form,  but,  since  the  demonstrations  have  in  no  way  been 
dependent  upon  the  signs  of  p  and  q,  it  is  obvious  that  equations  under 
the  three  other  forms  enjoy  the  same  properties. 

The  fifth  property  enables  us  to  solve  an  equation  directly,  without 
completing  the  square  in  the  first  member  j  but  it  is  well  to  require 
beginners  to  complete  the  square,  until  familiar  with  the  principles. 


300  EQUAXrONS    OF    TUE    SECOND    DEGREE, 

GENERAL   EXAMPLES. 

1.  Solve  the  equation,  ax^  +  ex  =  ma. 
Ans. 


c  /  f2         „  c  /  C" 

2«        V  4a^  2rt        V  4a2 

or,  .r  = -^ — ,  x"  = ^—T . 

2a  la 

2.  Solve  the  equation,  Ix^  —  Ux  =  +  280. 

Ans.  a:'  =  1  +  v^ll^  ■^■"  =  1  —  n/'IT- 

3.  Solve  the  equation,  nx^  —  mx  =  nm. 


^  ,  111  +  y/m^  +  4n^m      ,,       m  y/m^  +  4«^?w 

^9(S.      X=    ■\-      -r ,     if       == . 

2n  2n 


4.  Solve  the  equation,  ax^  —  hx  =  mx 

A7 

5.  Solve  the  equation,  4.t^  —  5.r  =  7a- 


J.ns.  0^'  =  0,  a;    = . 


Ans.  x'  =  0,  a"  =  3. 

6.  Solve  the  equation,  nx^  —  mx  =  6x. 

Ans.  x'  =  0,  .c'  =  — ■ . 


How  might  the  answers  to  5  and  6  be  deduced  directly  from  the 
answer  to  4. 

7.   Solve  the  equation,  — x^ — px  =^  +  §'• 


8.  Solve  the  equation,  x^  -\-  ^x  —  #  =  0. 

Ans.  ^'  ~  +  3j  x"  =  —  3. 

9.  Solve  the  equation,  4.^'  +  lOx  —  6  =  0. 

Ans.  a:'  =  +  ^,  x"  =  —  3. 

10.  Solve  the  equation,  x^  —  |  a:  —  #  =  0. 

'  Ans.  x'  =  +  3,  x"  =  —  A. 


EQUATIONS    OF    THE    SECOND    DEGREE.  301 

11.  Solve  the  equation,  x^  +  x  -\-  -:^\  =  0. 

Ans.  ; 

12.  Solve  the  equation,  x^  —  x  +  -f^  =  0. 

Ans.  x'  =  +  |,  x"  =  +  |. 

13.  Solve  the  equation,  16:/;^  —  16x  +  3  =  0. 

Ans.  x'  =  +  I,  x"  =  +  f 

14.  Solve  the  equation,  x^  —  ax  +  ah  =  +  hx. 

Ans.  x'  =  +  a,  x"  =  +  h. 

15.  Solve  the  equation,  x^  +  ax  +  ab  =  —  bx. 

Alls,  x'  =  —  a,  x"  =  —  b. 

16.  Solve  the  equation,  mx'  -\ m  =  -——. 

i  ,         ^        n  ^ 

Ans.  X  =  -J-,  X    = . 

0  a 

^_    _,  ,       ,  .  „      vibx  max 

li.  Solve  the  equation,  mx' vi  = ; — . 

'  «  b 

4  ,  ^  ,i  ^ 

Ans.  X  =  +  —,  X    = ^. 

a  b 

__    „  ,       .  .  ,       rt.r        mx  px 

18.  Solve  the  equation,  x^ -. =  —  ex  — ^—. 

b  n  q 

Ans.  x'  =  0,  x"  =  ---{ c  —  — . 

b        n  q 

-„     r,  ,       ,  .  ,    ,  ax       mx  px 

19.  Solve  the  equation,  «  +  -y-  +  —  =  ex  +  ~. 

'■  b  H  q 

A  inn  pa         m 

Ans.  .r'  =  0,  x"  =  c  +  =^ . 

q         b         n 

20.  Solve  the  equation,  x^  +  mx  +  ex  +  \/n  x-{-  m\r  =  0. 

Ans.  x'  =  0,  cc"  =  —  m  —  c  —  y/n  —  «t^. 

21.  Solve  the  equation,  x^  —  mx  —  ex  —  ^n  x —  m^x  =  0. 

Ans.  x'  =  0,  x"  =  m  +  c  +  ^n  +  m^. 

These  equations  show  that,  when  the  sign  of  the  coefficient  of  x  has 
been  changed,  and  the  other  terms  left  unaltered,  the  values  will  be 
numerically  the  same,  but  affected  with  contrary  signs.     They  show 
26 


802  EQUATIONS    OF    THE    SECOND    DEGREE. 

also  tliat,  when  all  the  terms  after  x^  are  negative,  the  second  member 
being  zero,  the  values  will  be  both  positive ;  and  when  the  terms  are 
•all  positive,  the  second  member  being  zero,  the  values  will  be  both 
negative. 

22.   Solve  the  equation,  mx^  +  px  =  —  q. 


Ans.   x'  =   — P+  ^p'  —  Amq    ^„  ^    —p—^p^  —  4mq 


"What  will  these  values  become  when  p^  =  Amq  ?  What,  when 
/  <  ^mq  ? 

23.  Solve  the  equation,  4:?^  ^  24a-  =  —  36. 

Ans.  x'  =  —  S,  x"  =  —  3. 

24.  Solve  the  equation,  4x^  —  24:r  =  —  36. 

Ans.  a-'  =  +  3,  x"  =  +  3. 

25.  Solve  the  equation,  4x^  —  24.r  =  —40. 

Ans.  .x'  =  +  3  +  ^—1,  x"=  +  3  —  ^— 1. 

The  following  important  consequences  flow  fi-om  the  last  four  ex- 
amples. 

Whenever  the  square  of  the  coefficient  of  x  is  not  greater  than  four 
times  the  product  of  the  coefficient  of  x"^  into  the  absolute  term,  the 
values  will  be  imaginary,  if  the  sign  of  the  absolute  term  is  negative. 
And,  whenever  the  square  of  the  coefficient  of  x  is  equal  to  this  nega- 
tive product,  the  two  values  of  x  will  become  identical. 

344.  Some  equations  may  be  treated  either  as  complete  or  in- 
complete. 

x^ 

1.   Solve  the  equation, -,  =  n. 

(a  —  xj 

Ans.   X  =  ^,  X    = . 


I  +  ^n  \  —  s/n 

For,  expanding  the  denominator  and  clearing  of  fractions,  we  have, 

a-2  —  n  (a^  —  2ax  +  x^),  or  (1  —  n)  x^  +  2anx  =  na^,  or  x^  + 

2an  na^         ^^  ,  an       .         /  ncv^  'a^ 

:- =  -z .      Hence,  :r  =  —  —- h  \  /  ^ ■  H = 

1— H        1  —  n  \  —  n         V   l—n^(l  —  ',if 

—  an  4-  ^na?  (1 — n)  +  ahi^ — an  -f  a^n  _  a^/n  (1  —  ,yn) 

1  —  n  1 — n  1  —  n  ' 

Now,  multiply  numerator  and  denominator  by  1  -f  ^/n,  and  we  have, 


EQUATIONS    OF    THE    SECOND    DEGREE.  303 

a^n(l  —  n)^  =      ''^"     .     In  the  same  way,  x"  cau  be  shown 
(1  — ?0(l  +  \/w)        1+  n/" 

equal  to =.     These  values  for  x'  and  x"  are  identical  with 

^  1— v/n 

those  before  obtained,  when  the  equation, ^  =  »,  was  treated  as 

^  (rt  —  x)2 

an  incomplete  equation. 

x^ 

2.  Solve  the  equation, :,  =  m^,  as  a  complete  equation. 

(rt  +  xy 

,           ,           am         ,,        — am 
Ans.  X  =  ;. — ■ ,  x    = 


X 

Solve  the  equation, :=-  =  q,  as  a  complete  equation 

(x  —  ^pf 

Ans.  x'  = 


1  -f  m'  1  4-  m 

a  complete  equation. 

s/pq         ,,  s/pq 


^q  —  1  y/q  +  l 

345.  In  the  foregoing  examples,  the  unknown  quantity  has  been 
freed  from  radicals.  Whenever  it  is  connected  with  radicals,  it  must 
be  freed  from  them  by  the  preceding  rules.,  and  then  the  process  will 
be  the  same  as  in  the  examples  already  given. 


1.  Solve  the  equation,  y/mx^+  nx  =  y/p 


n  4-  v/?i^  +  4m»      „             n  — ^n^  +  4mp 
^"^-  •'^  = ^ '  '■    = 2^1 • 

2.  Solve  the   equation,   V^V  +  nx  =  p,  as  an  equation  of  the 
second  degree.  .  ,  _       j^  n — P 

~  m  +  ?i'  m  —  n 

3.  Solve  the  equation,  ^mx^  +  nx  =  p,  as  an  equation  of  the 

second  degree.  ,  ,  P  tr  V 

Ans.  x  — — == ,x    = =r . 

•J  m  -{-  n  y/m       " 


4.  Solve  the  equation,  ^If  +    ^  "^  "^    =Sh 

'^r+x         Vl+.r 

A7}s.  a:'  =  +  8,  x"=—^. 

5.  Solve  the  equation, ( —=.  =  m. 

«  +  '-«  ^/a  +  x  

Aus.   X  ^^ ■ .  X  ^^^ ^r 

What  will  these  values  become  when  m  =  2?     What,  when  m  <^ 


9? 


oO-i       EQUATIONS  OF  THE  SECOND  DEGREE. 

6.  Solve  the  equation,  -5-— h  ■ —       "      =  -• 

Ans.  x'  =  —  2,  a;" 

^  ,       ,  •        n/1  +  .c         1  +  j- 

7.  Solve  the  equation,  -r; — ;— — 1 =  1 


J.JIS.  a;' 


l-\-x         ^1  + 
— 3+v/— 3     „       — 3  — v^^ITs 


a;  +  a  +  c 

8.  Solve  the  equation,  =  =  1. 

■,/2cx  -\-  4ac 


Ans.  x'  =  — a  +  ^/2ac  —  c^,  x"  =  —  a —  v/2ac  —  c\ 
"What  do  these  values  become  when  c^  =  2ac ?     What,  ■when  c'^2aP 

^  -|_  2  +  4 
9.  Solve  the  equation,  —  =z  1. 


s/Sx  +32 


Ans.  x' 


10.  Solve  the  equation 


+  2  +  5 


v/10^  +  40 
^/!s.  x'  =  —  2  +  ^/^^^,  x" 


v/— 5. 


11.   Solve  the  equation,  -J  x^  -\-  hx  =^  y/  ax  -{■  y/bx. 

Ans.  x!  =  0,  a;"  =  a  +  2  s/  ah. 
Verify  all  the  preceding  results. 


DISCUSSION    OF    COMPLETE    EQUATIONS    OF    THE    SECOND 
DEGREE. 

346.  The  four  forms  of  these  equations  are 

3? — px  =  q, 
X?  +  px  =  —  2', 
0?  —  px  z=  —  q, 
and  the  corresponding  values. 


First  form. 


EQUATIONS  OF  THE  SECOND  DEGREE. 


305 


='=f+\/^+^' 

="=f-\/^+. 

— f^\/?- 

--f-\/^-^> 

'=f-\/?- 

f-\/^-^. 


Second  form. 


Third  form. 


Fourth  form. 


Now,  we  observe  that  the  first  and  second  forms  diflfcr  only  in  the 
sign  of  X,  and  that  x'  in  the  first  form  is  the  same  as  x"  in  the  second, 
taken  with  a  contrary  sign ;  x"  in  the  firet  form  is  the  same  as  x'  in 
the  second,  taken  with  a  contrary  sign.  A  similar  remark  may  be  made 
in  regard  to  the  values  of  the  third  and  fourth  fonus.  We  conclude, 
then,  that  to  change  the  signs  of  the  values,  without  altering  them  nu- 
merically, we  have  only  to  change  the  sign  of  the  coofilacient  of  the  first 
power  of  X.  Thus,  x^  +  .x  =  2,  when  solved,  gives  x'  =  1,  x"  =  —  2; 
and  x^  —  .X  =  2,  gives  x'  =  2,  x"  =  —  1 . 

In  like  manner,  x^  —  5x  =  4,  when  solved,  gives  the  two  values, 
x'  =  +  4,    .x"  =  +  2 ;    and   x^  +  5.r  =  —  4,   when  solved,  gives 


If  we  proceed  to  extract  the  scjuarc  root  of  V+g', 


in  the  two  values 


of  the  first  form,  we  will  get  jr-,  plus  a  series  of  other  terms.     Hence, 


the  two  values  will  become  x  = 


+  4-  +  other  terms ;  and  x" 


P        P 
—  ^ ^ other  terms. 

Hence,  the  first  value  is  positive  and  the  second  negative,  and  the 
second  is  numerically  greater  than  the  first,  because  the  radical  is  added 
to  the  quantity  without  the  radical  in  the  second  value,  and  subtracted 
from  it  in  the  first  value. 

26*  u 


306       EQUATIONS  OF  THE  SECOND  DEGREE. 

By  extracting  tlie  square  root  of  "^  +  f?    approximatively  in    the 

vaJucs  of  tlic  second  form,  we  would  see  that  the  first  value  was  positive 
and  the  second  negative,  and  that  the  second  was  numerically  greater 
than  the  first.  We  have  anticipated  this,  from  what  has  been  said 
before,  as  to  the  interchange  of  position  and  the  change  of  sign  between 
the  values  of  the  first  and  second  form. 

In   the  third  form  both  values  are  negative,  because  —  ^   is   ^ 


\ 


If 

/  -J  —  q.     The  second  value  is  numerically  greater  than  the  first. 

In  the  fourth  form  both  values  are  positive,  and  the  first  is  numeri- 
cally greater  than  the  second.  These  results  might  have  been  antici- 
pated from  our  knowledge  of  those  in  the  third  form. 

347.  The  discussion  of  the  signs  of  the  values  in  the  four  forms 
may  be  made  otherwise,  thus :  we  know,  from  the  fourth  property  of 
equations  of  the  second  degree,  that  th.e  product  of  the  values  in  the 
first  form  must  be  equal  to  —  q.  Hence,  the  values  must  have  con- 
trary signs.  And  we  know,  from  the  third  property,  that  the  sum  must 
be  equal  to  —  p.  Hence,  the  negative  value  is  the  greater.  For, 
when  we  add  a  negative  and  positive  quantity  together,  and  their  sum 
is  negative,  we  know  that  the  negative  quantity  is  greater  than  the 
positive. 

For  the  second  form,  the  product  of  the  values  is  equal  to  —  q. 
Hence,  the  values  have  difi'erent  signs.  Their  product  is  equal  to  4-  p. 
Hence,  the  positive  value  is  greater  than  the  negative. 

For  the  third  form,  the  product  of  the  values  is  equal  to  -|-  q.  Hence, 
the  values  are  both  positive,  or  both  negative.  But  their  sum  is  equal 
to — p.  Hence,  they  are  both  negative.  "We  cannot  decide  from  this, 
however,  which  is  the  greater,  the  first  or  second  value. 

For  the  fourth  form,  the  product  of  the  values  is  equal  to  -\-  q. 
Hence,  the  values  must  have  like  signs,  and  must  be  both  positive,  or 
both  negative.  But  their  sum  is  positive,  being  equal  to  -j-  p  ;  hence, 
they  are  both  positive.  We  cannot  decide  from  this,  however,  which  is 
the  greater,  the  first  or  the  second  value. 


EQUATIONS    OF    THE    SECOND    DEGREE.  30' 


IRRATIONAL,    IMAGINARY,   AND  EQUAL  VALUES. 

348.  The  values,  in  all  tlie  forms,  will  be  irrational  whenever  the 
root  of  the  radical  cannot  be  extracted  exactly.  In  that  case,  the  ap- 
proximate value  of  the  radical  can  be  determined  to  within  a  vulgar  or 
decimal  fraction,  and  then  the  approximate  values  of  the  unknown 
quantity  will  be  known. 

There  can  be  no  imaginary  values  in  the  first  two  forms,  because 
the  quantities  under  the  radical  are  affected  with  positive  signs.  "When- 
ever, then,  the  absolute  term  of  an  equation  is  positive,  we  know,  cer- 
tainly, that  there  are  no  imaginary  values  in  the  equation. 

The  third  and  fourth  forms  contain  imaginary  values,  only  when  q  is 

f?  . 

^  ^.     For,  in  that  case  only,  is  the  prevailing  sign  under  the  radical 

negative.  Whenever,  then,  an  equation  is  put  under  the  third  or  fourth 
form  (the  unknown  quantity  being  freed  from  radicals,  and  the  coeflS- 
cient  of  x^  being  made  plus  unity),  we  can  readily  tell  whether  there  are 
imaginary  values.  "We  have  only  to  square  half  the  coefficient  of  the 
first  power  of  x,  and  compare  the  result  with  the  absolute  term. 

The  equations,  x^  -|-  lOx  =  —  30,  and  a?  —  lOx  =  —  30,  both 
contain  imaginary  values,  because  (5)*  <^  30.  Imaginary  values,  being 
always  similar  in  form,  are  said  to  enter  the  equation  in  pairs. 

Since  the  two  values  of  each  of  the  four  forms  differ  only  by  the 
radical,  it  is  obvious  that  the  disappearance  of  the  radical  will  cause 
these  values  to  become  identical.  But,  the  radical  cannot  be  made  to 
vanish  in  the  first  and  second  forms,  because  the  quantities  under  the 
sign  are  positive.     It  will  vanish,  however,  in  the  third  and  fourth 

forms,  whenever  ^=zq.     The  values  of  the  third  form  will  then  both 

become — i^,  and  of  the  fourth   form  +  ^.     The    equality  of  the 

values,  in  the  third  and  fourth  forms,  differs  from  the  equality  of  the 
values  of  incomplete  equations.  In  the  latter,  it  is  equality  only  in  a 
numerical  sense;  in  the  third  and  fourth  forms,  absolute  equality. 

If  we  substitute  ^  for  q  in  the  third  and  fourth  forms,  the  equations 
will  become  x^  -}-  px  =  —  -^,  and  x^ — px  =  —  ^,  or,  by  transposi- 


tion 


,  x^  +  px  +  ^  =  0,  and  x^  — px  +  ^  = 


308       EQUATIONS  OF  THE  SECOND  DEGREE. 

The  first  member  of  one  equation  is  the  square  of  (x  +  ^),  and  of 

the  other  (x  —  ^).     Hence,  when  there  are  equal  values  in  the  third 

and  fourth  forms,  the  first  members  of  those  equations  will  be  perfect 
squares,  provided  the  second  members  are  zero. 

The  following  are  illustrations  :  x^  +  Gx  =  —  9,  and  cc^  —  6x  = 
—  9;  x^  +  Sx  =  —lQ,andx^  —  Sx  =  —  lQ. 


SUPPOSITIONS  MADE   UPON  THE   CONSTANTS. 

349.  Known  terms  are  frequently  called  constants,  and  unknown 
terms,  variables.  In  the  equation,  x^  -\- px  =  q,  p  and  q  are  the  con- 
stants, and  x  the  variable. 

Let  us  make  the  constant  q  —  o,  in  the  four  forms.  By  going  back 
to  the  solved  equations,  we  see  that  the  values  will  become 


^"  _     p     p 

2""  2  ~ 

2h 

First  form. 

^■  =  f  +  f=., 

Second  form 

^'  =  -f  +  f  =  o, 

P, 

Third  form. 

Fourth  form 

We  see  that  the  first  and  third  forms  have  the  same  values,  and  that 
the  second  and  fourth  forms  have  the  same  values.  This  ought  to  be 
so,  for  the  hypothesis,  g-  =  0,  makes  the  first  and  third  equations  iden- 
tical, and  also  makes  the  second  and  fourth  equations  identical. 


EQUATIONS    OF    THE    SECOND    DEGREE.  300 

We  have  gotten  the  foregoing  series  of  values  by  making  (j'  =  0  iu 
the  solved  equations.  Ought  we  not  to  get  the  same  results  by  opera- 
ting upon  the  given  equations  themselves  ?  When  3  =  0,  the  first 
and  third  forms  become  x^  -{-  px  =  0,  ov  x  (x  -{-  p)  :^  0 ;  an  equation 
which  can  be  satisfied  by  placing  either  factor  equal  to  zero.  Hence, 
X  =  0,  and  X  +  p  =  0 ;  or,  a;'  =  0,  and  x"  =  —  p ;  the  same  as 
before  obtained.  The  second  and  fourth  forms  become,  when  q  =  ^, 
x^  — 2)x  =  0,  or  X  (x  — p)  =  0.  Hence,  x"  =  0,  and  x'  =z  +  p,  as 
before. 

When  2>  =  0,  we  have  this  series  of  values  : 

c'  =  -\-\/q,  x"  =z  — v/gr,  in  the  first  and  third  forms. 

x'  =  4-v/ —  'j>  •^■"  =  — \/ —  'J,  in  the  second  and  fourth  forms. 

These  results  can  be  obtained  directly  from  the  equations ;  for  the 
first  and  second  become  x^  =  q,  and  the  third  and  fourth,  x^  =  —  q. 
These  two  equations  will  plainly  give  the  four  preceding  values. 

When  p  =  0,  and  q  =  0,  the  system  of  values  reduces  to  x'  =  +  0, 
x"  =  +  0,  in  each  of  the  forms.  This  ought  to  be  so ;  for,  in  that 
case,  the  four  equations  reduce  to  the  same,  x^  =  0. 

The  solutions  of  the  four  equations  give  formulas  which  can  be  ap- 
plied to  particular  examples.     Thus,  the  values  of  the  first  form   are 


Let  it  be 


-'=-f  +  \/f  +  q,  and  a^"  =  -J^-syj^  + 

required,  now,  to  apply  these  results  as  formulas  to  find  the  values  of 
the  equation,  x^  +  x  =  2.  Then,  jj  =  1,  and  q  =  2.  Hence,  x'  z= 
—  h  +  v/i-  +  2=  —^.^3^1.  And  x"  =  —i  —  ^i-^2  =  — 
k  —  #=  —  2. 

We  have  used  as  formulas  the  values  x'  and  x",  belonging  to  the 
first  form,  because  the  equation,  x^  +  x  =  2,\s  of  that  form.  And,  of 
course,  we  must  always  apply  as  formulas  the  values  found  in  the  form 
to  which  the  given  equation  belongs. 


EXPLANATION  OF  LAIAGINARY  VALUES. 

350.  We  have  seen  that  the  values  of  the  third  and  fourth  forms  be- 

come  imaginary  when  q  was  made  greater  than  ^.     It  remains  to  ex- 

plain  the  cause  of  the  imaginary  results,  and  to  ascertain  what  they  mean. 

In  the  third  form,  the  product  of  the  values  is  equal  to  +  q,  and 

their  sum  equal  to  — p.     The  values,  then,  are  both  negative;  and 


310  EQUATIONS    or    THE     SECOND    DEGREE. 

we  liave  previously  seen  that  tliey  were  unequal  wlicu  ^  was  unequal 
to  q.  If  tlicy  were  equal,  caeli  must  be  —  ^.  Let  x  represent  the 
execss  of  the  greater  over  — ^.  Then  the  value  which  is  numerically 
the  greater  will  be  represented  by  —  ( {y  +  x\,  and  the  smaller  nu- 
merically will  be  represented  by  —  ("^ x\.      Their   product   will 

be  ^^ x^.     Now,  it  is  evident  that  this  product  will  be  the  greatest 

possible,  when  x  =  0.  This  condition  makes  the  two  values  equal,  and 
makes  their  greatest  product  ^ .     But  q  represents  their  product,  and 

it,  therefore,  can  never  be  greater  than  -^.       Hence,   when  we  make 

4 

-7-  ^  <?,  we  impose  an  impossible  condition.     An  imaginary  solution, 

then,  indicates  an  impossible  condition. 

Ought  not  the  equation  to  point  out  an  absurdity  when  an  impossi- 
ble condition  is  imposed  ?     When  the  square  is  completed  in  the  third 

form,  the  equation  \s,  x^  -\-  px  -\    ^  =  — q,  ox  (x  -\-  ^\  ==j  —  q. 

p- 
Now,  when  q"^  ^,  the  second  member  is  essentially  negative,  whilst 

the  first,  being  a  square,  is  essentially  positive.  We  then  have  a  nega- 
tive quantity  equal  to  a  positive  quantity,  which  is  absurd.  A  careful 
inspection,  then,  of  the  equation  which  j^roduces  imaginarij  values  loill 
show  an  ahsurdity. 

We  have  examined  only  the  third  form,  but  the  preceding  conse- 
quences may  be  readily  deduced  by  an  examination  of  the  fourth  form. 

The  following  problem  will  illustrate  more  fully  the  subject  of  ima- 
ginary values. 

Required  to  divide  the  number  10  into  two  parts ;  such,  that  their 
product  will  be  equal  to  30. 

Let  X  be  one  of  the  parts,  then  10  —  x  will  be  the  other,  and,  from 
the  conditions  we  get  x  (10  —  a;)  =  30,  or  —  x^  -\-  l^x  =  30.  Mul- 
tiplying both  members  by  minus  unity,  we  get  x^  —  lOx  =  — 80,  and, 
completing  the  square,  x^  —  10a;  -f  25  =  —  5.  Hence,  x'  =  5  -f 
^/— 5,  and  x"  =  5  s/^^ 


GENERAL    PROBLEMS.  311 

The  values  are  imaginary,  as  they  ought  to  be ;  for  the  greatest  pro- 
duct that  we  can  form  of  two  numbers,  whose  sum  is  10,  is  25. 
Hence,  we  have  imposed  an  impossible  condition,  and  the  imaginary 
values  point  out  this  impossibility.  The  equation,  x^  —  lOx  -|-  25  = 
—  5,  may  be  written  (x  —  5)-  =  —  5 ;  that  is,  a  positive  quantity 
equal  to  a  negative  one,  which  is  absurd.  The  imaginary  values  will, 
however,  satisfy  the  given  equation,  as  they  ought  to  do,  since  they 
have  been  derived  from  it. 


EXPLANATION  OF  NEGATIVE  SOLUTIONS. 

351.  Required  to  find  a  number  whose  square,  augmented  by  three 
times  the  number,  and  also  by  4,  shall  be  equal  to  2. 

The  equation  of  the  problem  is  a;'^  +  3x  +  4  =  2. 

This  equation,  plainly,  cannot  be  satisfied  in  an  arithmetical  sense ; 
for  4  is  already  greater  than  2,  and  must  be  still  greater  when  aug- 
mented by  y^  and  3x.  Solving  the  equation,  we  get  x'  ^  —  1,  and 
:c"=  — 2. 

These  negative  values  satisfy  the  equation,  but  do  not  fulfil  the  con- 
ditions of  the  enunciation.  For,  x  being  negative,  the  equation  will 
become  x-  —  3a;  -t-  4  =  2.  The  true  enunciation  of  the  problem,  then, 
is  :  Required  to  find  a  number,  whose  square,  diminished  by  three  times 
the  number,  and  that  remainder  increased  by  4  will  give  a  result  equal 
to  2.  The  equation  of  this  problem,  when  solved,  will  give  the  two 
positive  values,  1  and  2. 

Negative  values  then,  in  equations  of  the  second  as  well  as  of  the 
first  degree,  satisfy  the  equation  of  the  problem,  but  do  not  fulfil  the 
enunciated  conditions ;  and,  since  negative  quantities  indicate  a  change 
of  direction  or  character,  those  negative  values  point  out  the  change 
that  must  take  place  in  the  enunciation,  in  order  that  its  conditions 
may  be  complied  with. 

GENERAL  PROBLEMS  INVOLVING  COMPLETE  AND  INCOM- 
PLETE EQUATIONS  OF  THE  SECOND  DEGREE. 

352.  1.  Required  to  divide  the  number  n  into  two  parts,  the  product 
of  which  shall  be  equal  to  m. 

Ans.  x'  =r  2  7i,  +  v^i?r  —  m,  x"  =  in  —  y/i)!"  —  in. 

What  will  these  values  become  when  m  =  W  ?  What,  when 
m  >  In"-  ?     Why  ? 


812  GENERAL    PROBLEMS    INVOLVING 

2.  Required  to  divide  the  number  10  into  two  sucli  parts  that  their 
product  shall  be  equal  to  25.  Ans.  x'  =  5,  x"  =  5. 

Is  the  disappearance  of  the  radical  always  connected  with  maximum 
values  ? 

3.  Required  to  divide  the  number  10  into  two  such  parts  that  their 
product  shall  be  equal  to  2G. 

Ans.  x'  =  b  -\-  >/— 1,  x"  =  5—  V — 1. 

Why  are  these  values  imaginary  ?     Verify  them  by  substitution  in 
the  equation  of  the  problem. 

4.  The  sum  of  two  numbers  is  a,  and  the  sum  of  their  squares  is  h, 
what  are  the  numbers  ? 


,       a  +  ^-Ih  —  a^      „        a  —  ^2b  —  a^ 

A7IS.    X    z=:  -: ,    X     =  ^ . 

When  do  these  values  become  imaginary  ?  when  equal  ?  What  is 
the  least  value  that  the  sum  of  the  squares  can  have  ? 

5.  The  sum  of  two  numbers  is  10,  and  the  sum  of  their  squares  is 
52,  what  are  the  numbers  ?  Aris.  x'  =  G,  x"  =  4. 

G.  The  sum  of  two  numbers  is  10,  and  the  sum  of  their  squares  is 
50,  what  are  the  numbers  ?  Ans.  x'  r=  5,  cc"  :=  5. 

7.  The  sum  of  two  numbers  is  10,  and  the  sum  of  their  squares  is 
40,  what  are  the  numbers  ? 

Ans.  x'  =  5  +  J^^,  if"  r=  5  —  -J — 5. 

How  may  the  values,  in  the  last  three  examples,  be  deduced  directly 
from  the  general  values  in  example  4  ? 

8.  A  Yankee  pedlar  bought  a  certain  number  of  clocks,  which  he 
sold  again  for  m  dollars.  His  gain  per  cent,  on  his  investment  was 
expressed  by  the  number  of  dollars  in  it.  How  much  did  he  pay  for 
the  clocks  ? 


Ans.  a;'  =  —  50  +  v/(25  +  m)100,  a;"  =  —  50  —  V(25  +  m)100. 

9.  Same  problem  as  the  last,  except  that  the  pedlar  sold  his  clocks 
for  375  dollars,  instead  of  m  dollars. 

Ans.  x'  =  150,  x"  =  —  250. 

What  is  the  meaning  of  the  negative  value? 


EQUATIONS.  OF  THE  SECOND  DEGREE.       313 

10.  A  number  of  partners  in  business  owe  a  debt  of  m  dollars,  but, 
in  consequence  of  the  failure  of  one  of  their  number,  each  of  the  solvent 
partners  has  to  pay  n  dollars  more  than  his  proper  proportion  of  the 
debt.     How  many  partners  were  there. 


A        -'  —  V     n  „  _  V     n 


11.  Same  problem  as  last,  except  that  the  debt  was  728  dollars,  and 
that  the  increased  portion  of  the  solvent  partners  was  60  j  dollars. 

Ans.  x'  =  4,  x"  z=  —  3. 

The  negative  value  is  readily  explained.  An  examination  of  the 
equation  of  the  problem,  after  —  x  has  taken  the  place  of  +  x,  wDl 
show  that  there  have  been  two  changes  of  condition,  and  the  correspond- 
ing enunciation  will  be  :  "  Several  partners  in  trade  owed  a  debt  of  728 
dollars,  and,  by  the  accession  of  another  partner  to  share  the  debt  with 
them,  their  individual  liability  was  diminished  by  GOf ;  required  the 
number  of  partners."  These  changes  in  the  enunciation  give  -f-  3  for 
the  number  of  partners,  and  the  result  can  be  verified.  For,  the  share 
of  each  in  the  debt  is,  242 f  dollars  before  a  new  partner  was  added  to 
the  firm,  and  but  182  dollars  afterwards.  And,  in  general,  for  problems 
involving  equations  of  the  second  degree,  there  must  be  two  changes 
in  the  enunciation  to  convert  a  negative  into  a  positive  solution. 

12.  The  difference  between  the  cube  and  the  square  of  a  number  is 
equal  to  twice  the  number.     Required  the  number. 

Ans.  x'  =  -\-  2,  x"  ^  —  1. 

13.  Required  to  .divide  a  quantity,  a,  into  (wo  such  parts,  that  the 
greater  part  shall  be  a  mean  proportional  between  the  whole  quantity 
and  the  smaller  part.     What  are  the  two  parts  ? 

Am.  Greater  part,  either ,  or —^ — ^— ;  and  lesser 

part,  either  |a  —  la  ^5,  or  -S«  +  la  \/b. 

Both  values  will  satisfy  the  enunciation,  in  one  sense ;  for,  a(%a  — 

Because,  by  performing  the  indicated  operations,  we  have  for  the  first 

27 


314  GENERAL    PROBLEMS    INVOLVING 

((2  ^2      g^2 

value,  2a-  —  .UV^  =  ^  —  -5- v/5  +  —  r=  -Ja"  —  ^"V"^J  and,  for 
tlie  second  value,  §a«  +  Ws/^  =  ^  -\-  W^b  + -j- =  2"'  +  iaVST 

The  second  value,  however,  does  not  satisfy  the  conditions  of  the 
problem,  understood  in  a  literal  sense,  for  the  greater  part, 

— v/^j  is  less  than  the  corresponding  smaller  part,  %a  +  -^\/~^,  and 

this  corresponding  part  is  greater  than  the  whole  quantity,  a.  The 
explanation  of  the  negative  solution  is  simple,  when  we  return  to  the 
equation  of  the  problem,  x^  =  a  (a  —  x).  When  x  is  negative,  this 
equation  becomes,  x^  =  a(a  -{•  x),  and  the  corresponding  enunciation 
ought  to  be  required  to  find  a  number,  which  shall  be  a  mean  between 
the  whole  quantity,  a,  and  the  sum  of  a  and  this  number. 

The  given  quantity,  a,  might  have  been  represented  by  a  straight  line, 

-f J-,  and  the  problem  then  would  have 

been  to  find  a  part,  BC,  which  should  be  a  mean  between  the  whole,  BA, 
and  the  part,  AC,  left  after  BC  was  taken  from  it.  Now,  when  the  ex- 
pression for  BC  became  negative  in  the  solution  of  the  problem,  it  in- 
dicated that  BC  must  be  laid  oif  on  the  left  of  the  point  B,  because 
this  distance  was  laid  off  on  the  right  of  B,  when  its  expression  was 

positive.     The  diagram  becomes -, 

and  AC  is  greater  than  AB.     This  agrees  with  the  second  value  of  AC, 

3  a 

—  a  +  7^  v/^  which  is  greater  than  a  or  AB.  The  negative  solu- 
tion here  indicates  then  a  change  of  direction,  and  the  equation  being 
of  the  second  degree,  there  are  two  corresponding  changes  of  condition. 
The  first  is  expressed  by  the  unknown  distance  being  sought  upon  the 
prolongation  of  AB,  instead  of  upon  AB  itself;  the  second  is  expressed 
by  the  unknown  distance  being  a  mean  between  AB,  and  A'B  j'lus  BC, 
instead  of  between  AB  and  AB  minus  BC. 

The  explanation  of  a  negative  solution  need  never  be  difiicult;  we 
have  only  to  change  +  x  into  —  a;  in  the  equation  of  the  problem,  and^ 
then  examine  and  see  what  the  resulting  equation  means. 

Some  of  the  following  problems  will  involve  one  of  the  four  methods 
of  elimination.     Whenever  the  equations  between  which  the  elimina- 


EQUATIONS    OF    TUE    SECOND    DEGREE-:  olo 

tion  is  to  be  effected  is  of  a  higher  degree  than  the  first,  the  method  of 
elimination  by  the  greatest  common  divisor  ought  to  be  employed. 

14.  Find  two  numbers  whose  sum  and  product  are  both  equal  to  a. 

.          ,        a  +  ^o?  —  4a      ,,        a  —  ^/a^  —  4« 
Ans.  X  = :; ,  X    = ,;^ , 


,       a — v/a^  —  4a  ^/a- — 4a  +  a 

y  = 2 — '^  "" — 2 • 

When  will  the  two  numbers  be  equal  ?     "When  imaginary  ? 

15.  Find  two  numbers  whose  sum  and  product  shall  be  equal  to  4. 

Ans.  2  and  2. 

16.  Find  two  numbers  whose  sum  and  product  shall  be  equal  to  2. 
Ans.  x'  =  1  +  y/—l,  x"  =1  —  >/^^^,  y'  =  l  —  y/'-^,  xf'  = 

1  +  ^/^=^. 

17.  Find  two  numbers  whose  sum  and  product  shall  be  both  equal 
to  5. 

,       5  +  v/5"    ,,       5  —  ^o~   ,      5  —  v^o"   ,,      5  +  Jb 
Ans.  x'  = ^ >  *■    = 2 '  ^  ^ 2 '  ^   ^ 2 —  " 

Do  the  values  in  problems  14,  16  and  17,  indicate  that  there  are  two 
distinct  sets  of  numbers,  or  that  the  second  set  is  the  same  as  the  first, 
differing  only  in  the  position  of  the  numbers. 

18.  Two  capitalists,  A  and  B,  invested  different  sums  in  trade.  A 
invested  half  as  much  as  B,  and  kept  his  money  in  trade  6  months, 
and  gained  one-twentieth  of  his  investment.  B  kept  double  the  amount 
that  A  had,  in  trade  for  9  months,  and  gained  §100  more  than  A. 
Supposing  that  their  respective  gains  were  proportional  to  their  re- 
spective capitals  and  the  periods  of  investment,  what  were  their 
capitals  ?  Ans.  A's  01000,  B's  82000. 

Why  was  the  second  value  of  x  rejected  ?  What  was  the  gain  per 
cent  of  A  and  B? 

19.  The  difference  between  two  numbers  is  a,  and  the  difference  be- 
tween their  cubes  is  h  ;  what  are  the  numbers  ? 


Ans 


•"  =  2-+\/-T2^-^    =Y-VT2^' 

^=-2-  +  \/-T2^'^    ==-Y-\/-l2^- 
When  will  x"  and  y"  be  real,  but  negative  ? 


316  GENERAL    PROBLEMS    INVOLVING 

AVhat  do  the  negative  values  satisfy  ?  When  will  the  two  values  of 
X  be  equal  ?  How  will  the  two  values  of  y  be  in  that  case  ?  When 
will  the  solutions  be  indeterminate?  How  many  conditions  must  be 
imposed  ? 

20.  The  difference  of  two  numbers  is  2,  and  the  difference  of  their 
cubes  152  ;  what  are  the  numbers  ? 

Ans.  First  number,  +  6,  or  —  4  ;  second  number,  +  4,  or  —  6. 
How  are  the  negative  values  explained  ? 

21.  The  difference  of  two  numbers  is  4,  and  the  difference  of  their 
cubes  16;  what  are  the  numbers? 

Ans.  First  number,  +  2 ;  second  number,  —  2. 

22.  The  diffcrenee  of  two  numbers  is  4,  and  the  difference  of  their 
cubes  15 ;  what  are  the  numbers  ? 


-^W^ 


23.  The  difference  of  two  numbers  is  zero,  and  the  difference  of 
their  cubes  zero ;  what  are  the  numbers  ? 

A71S.  x'  and  x",  both  =  -—,  and  i/'  and  ^",  also  both  =  -— . 

24.  The  year  in  which  the  translation  began  of  what  is  called  King 
James'  Bible,  is  expressed  by  four  digits.  The  product  of  the  first, 
second  and  fourth,  is  42 ;  the  fourth  is  one  greater  than  the  second, 
and  the  sum  and  difference  of  the  first  and  third  are  both  equal  to  one. 
Required  the  year.  Ans.  1607. 

25.  The  year  in  which  Decatur  published  his  official  letter  from  New 
London,  stating  that  the  traitors  of  New  England  burned  blue  lights 
on  both  points  of  the  harbour  to  give  notice  to  the  British  of  his  at- 
tempt to  go  to  sea,  is  expressed  by  four  digits.  The  sum  of  the  first 
and  fourth  is  equal  to  half  the  second ;  the  first  and  third  are  equal  to 
each  other;  the  sum  of  the  first  and  second  is  equal  to  three  times  the 
fourth,  and  the  product  of  the  first  and  second  is  equal  to  8.  Ee- 
quired  the  year.  Ans.   1813. 

26.  The  year  in  which  the  Governors  of  Massachusetts  and  Connec- 
ticut sent  treasonable  messages  to  their  respective  Legislatures,  is  ex- 
pressed by  four  digits.     The  square  root  of  the  sum  of  the  first  and 


EQUATIONS    OF    THE    SECOND    DEGREE.  317 

second  is  equal  to  3 ;  the  square  root  of  the  product  of  the  second  and 
fourth  is  equal  to  4 ;  the  first  is  equal  to  the  third,  and  is  one-half  of 
the  fourth.     Required  the  year.  Ans.  1812. 

27.  A  gentleman  puts  out  a  certain  capital  at  an  interest  of  5  per 
cent. ;  the  product  of  the  interest  for  12  mouths  by  the  interest  for  one 
month,  is  just  one  fourth  of  the  principal.     What  is  the  capital  ? 

Ans.  a/  =  0,  x"  =  $1200. 

28.  Some  of  the  New  England  States  were  fully,  and  some  partially, 
represented  in  the  Hartford  Convention,  which,  in  the  year  1814,  gave 
aid  and  comfort  to  the  British  during  the  progress  of  the  war.  If  4  be 
added  to  the  number  of  States  fully  and  partially  represented,  and  the 
square  root  of  the  sum  be  taken,  the  result  will  be  the  number  of  States 
fully  represented;  but  if  11  be  added  to  the  sum  of  the  States  fully  and 
partially  represented,  and  the  square  root  of  the  sum  be  taken,  the 
result  will  be  equal  to  the  square  root  of  8  times  the  number  of  States 
partially  represented.  Required  the  number  of  States  fully  and 
partially  represented. 

Ans.  Three  fully  represented ;  two  partially  represented. 

29.  The  sum  of  two  numbers  is  a,  the  sum  of  their  squares  h,  and 
the  sum  of  their  cubes  c.     What  arc  the  numbers  ? 

A71S.  x' 


^+ 

.s/.+^^- 

-/>)      ,. 

2 

.J\/«= 

.,^ 

-ab) 

I 

V" 

.2      ,     4(C— «^) 

andf 

=  h 

i.\A= 

c 

-ab) 
I 

Have  we  two  distinct  sets  of  values,  or  but  one  set,  with  an  inter- 
change of  position  ?  When  will  the  values  of  x  and  y  be  equal  ?  When 
imaginary  ?     When  negative  ? 

30.  The  sum  of  two  numbers  is  7,  the  sum  of  their  squares  25,  and 
the  sum  of  their  cubes  91.     What  are  the  numbers  ? 

Ans.  x'  =  4,  a;"  =  3 ;   i/'  =  3,  //"  =  4. 

31.  The  sum  of  two  numbers  is  6,  the  sum  of  their  squares  is  18, 
and  the  sum  of  their  cubes  54.     What  are  the  numbers  ? 

Ans.   Both  3. 

32.  The  sum  of  two  numbers  is  5,  the  sum  of  their  squares  20,  and 
the  sum  of  their  cubes  50.     What  are  the  numbers  ? 


Ans.  x!  =  %  +K/— 15,  x"  =  4  —  \  ^—  15 ;  y  =  4  —  *  n/— 15, 

27* 


318  GENERAL  PROBLEMS  INVOLVING 

33.  Two  travellers  started  at  the  same  time  from  two  cities,  C  and 
W,  and  travelled  toward  each  other.  They  found,  on  meeting,  that  the 
traveller  from  C  had  travelled  150  miles  more  than  the  other  traveller, 
and  that  by  continuing  at  the  same  rate,  he  could  reach  W  in  five  days; 
whereas,  it  would  take  the  traveller  from  W  twenty  days  from  the  time 
of  meeting  to  reach  C.  Required  the  distance  between  W  and  C,  the 
rate  of  travel  per  day  of  the  two  travellers,  and  the  time  that  had 
elapsed  before  their  meeting. 

Ans.  Distance  between  W  and  C,  450  miles;  the  rates,  30  and  15 
miles  per  day;  the  time  elapsed  before  meeting,  10  days. 

34.  The  sum  of  three  numbers  is  15,  the  sum  of  their  squares  93, 
and  the  third  is  half  the  sum  of  the  first  and  second.  What  are  the 
numbers  ?  Ans.  8,  2  and  5. 

35.  A  man  bought  a  tract  of  land  for  811  per  acre,  and  sold  it  again 
at  a  less  price,  his  loss  per  cent,  on  the  sale  being  expressed  by  the  price 
per  acre  which  he  received.     What  did  he  sell  the  land  for  ? 

Ans.  10  dollars  per  acre. 

36.  In  the  year  1G37,  all  the  Pequod  Indians  that  survived  the 
slaughter  on  the  Mystic  River  were  either  banished  from  Connecticut, 
or  sold  into  slavery.  The  square  root  of  twice  the  number  of  survivors 
is  equal  to  jLth  that  number.    What  was  the  number  ?      Ans.  200. 

37.  A. Southern  Planter  bought  m  acres  of  cultivated,  and  as  many 
of  uncultivated,  land.  He  got  b  more  acres  of  uncultivated  than  of 
cultivated  land  per  dollar,  and  the  whole  cost  of  the  cultivated  exceeded 
that  of  the  uncultivated  by  c  dollars.  How  much  did  he  pay  per  acre 
for  each  kind  of  land  ? 

1 


Ans.  x'  =  -— ^    ,    1  .    /  i  (4m  +  be)  . 
1 


-i>  a       /b  (4m  -f  be )       for    cultivated    land ;    y   = 


^  _j.  .       /b  (im  +  be),  y"   =   '!__  y^  ./b  (4m  +  be), 

for  uncultivated  land. 

38.  A  Southern  Planter  purchased  100  acres  of  cultivated,  and  100 
acres  of  uncultivated  land,  the  former  costing  him  $500  more  than  the 


EQUATIONS    or    THE    SECOND    DEGREE.  319 

latter ;  the  smaller  cost  of  the  latter  resulted  from  his  getting  for  every 
dollar  y'^th  of  an  acre  more  of  the  uncultivated  land  than  of  the  cultivated. 
What  was  the  cost  per  acre  of  the  cultivated  and  uncultivated  land. 
Ans.  Former,  $10  per  acre;   latter,  $5  per  acre. 

Verification.  100  acres  at  $10  per  acre  will  cost  SIOOO,  and  100  acres 
at  $5  will  cost  $500.  A  dollar  will  buy  ^^ih  of  an  acre  of  the  culti- 
vated, and  ith  of  an  acre  of  the  uncultivated  land ;  and  the  difference 
between  ith  and  y^th  is  jgth.  So,  a  dollar  will  buy  -jjjih  of  an  acre 
more  of  the  uncultivated  than  of  the  cultivated  land. 

.39.  In  the  year  1853,  a  number  of  persons  in  New  England  and 
New  York,  were  sent  to  lunatic  asylums  in  consequence  of  the  Spiritual 
Eapping  delusion.  If  14  be  added  to  the  number  of  those  who  became 
insane,  and  the  square  root  of  the  sum  be  taken,  the  root  will  be  less 
than  the  number  by  42.     Required  the  number  of  victims. 

A71S.  50. 

Why  is  the  second  value  of  x  rejected  ? 

40.  Two  farmers,  A  and  B,  invest  each  a  certain  amount  in  the 
Central  Railroad  of  North  Carolina.  After  a  time,  A  sells  his  stock 
for  $150,  and  gains  as  much  per  cent,  on  his  outlay  as  B  invested. 
B  also  sells  his  stock' and  gets  S122-  more  than  he  gave  for  it,  but  his 
gain  per  cent,  on  his  outlay  is  only  half  as  great  as  that  of  A.  Required 
the  amount  invested  by  each.  Ans.  A,  $100;  B,  $50. 

Why  is  the  negative  solution  rejected? 

Verification.  50  per  cent,  on  $100  is  $50.  And  since  A  sold  his 
stock  for  $50  more  than  he  gave  for  it,  he  gained  50  per  cent,  on  the 
$100  of  outlay.  B  gained  25  per  cent  on  his  outlay,  and  25  per  cent, 
upon  $50,  is  $12^. 

41.  Two  travellers  set  out  at  the  same  time,  the  one  from  A,  and  the 

otlier  from  C,  and  travel  towards     -, ■ 

each  other  at  uniform  rates.  After  meeting  at  B,  the  traveller  from  A 
is  a  days  in  reaching  C,  and  the  traveller  from  C,  c  days  in  reaching  A. 
How  long  was  it  after  the  time  of  starting  until  they  met  at  B,  and 
how  long  was  each  traveller  in  performing  the  distance  A  C. 

Time  of  meeting,  dc  y^ac.  Traveller  from  A,  a  ±  x/ac;  traveller 
from  C,  c  ±  y/ac. 


320  a  EN  Ell  A  L     niOBLEMS     INVOLVING 

What  is  the  moaniug  of  the  negative  sign  in  the  values  of  the  time  ? 
When  only  can  the  time  occupied  by  the  traveller  from  A  be  negative  ? 
How  then  will  both  expressions  for  the  time  occupied  by  the  other 
traveller  be  affected,  with  the  positive  or  negative  sign  ? 

42.  Same  problem  as  the  last,  except  that  the  traveller  from  A  is  9 
days  between  the  points  B  and  C,  and  the  traveller  from  C,  25  days 
between  the  points  B  and  A. 

Ans.  Time  of  meeting,  ±  15  days ;  time  occupied  by  traveller  from 
A,  24,  or  —  6  days ;  time  occupied  by  the  other  traveller,  40,  or  —  10 
days. 

43.  In  a  certain  bank  there  are  $438  worth  of  5  and  3  cent  pieces, 
the  number  of  the  latter  is  exactly  the  square  of  the  number  of  the 
former.     Required  the  number  of  pieces  of  each  kind. 

Ans.  120  five  cent  pieces ;  14,400  three  cent  pieces. 

44.  Required  to  find  two  numbers,  such  that  their  sum,  their  pro- 
duct, and  the  difference  of  their  squares,  may  all  be  equal  to  each  other. 

Ans.  x'  =  %  +  ^J,  x"  =  ^-^~^',  y'  =  ^  +  ^J,  y  =  ^  — ^/"|: 

45.  In  the  year  1706  the  French  made  a  descent  upon  Charleston; 
but  "  South  Carolina,"  says  Bancroft,  "  gloriously  defended  her  territory, 
and,  with  very  little  loss,  repelled  the  invaders."  A  certain  number 
of  the  French  were  killed  and  wounded,  and  100  were  taken  prisoners. 
The  number  of  killed  and  wounded  was  to  the  number  of  uninjured, 
including  the  prisoners,  as  1  to  3.  And  the  square  of  the  number  that 
escaped  in  safety  from  the  expedition,  was  to  the  square  of  the  number 
killed  and  wounded,  as  6i  to  1.  Required  the  number  of  invaders, 
and  the  number  of  killed  and  wounded. 

Ans.  800  invaders,  and  200  killed  and  wounded. 
*    Verification.  If  200  were  killed  and  wounded,  then  600  were  un- 
injured, and  200  :  600  :  :  1  :  3.     And,  since  100  were  taken  prisoners, 
500  escaped  without  harm  from  the  expedition,  and  (500)^  :  (200)^  :  : 
6J  :1. 

Why  is  the  value  connected  with  the  negative  sign  of  the  radical 
rejected? 

46.  In  the  year  1842  South  Carolina  converted  the  citadel  at 
Charleston,  and  the  magazine  at  Columbia,  into  military  academies, 
which  were  to  be  supported  by  the  sum  of  money  appropriated  annually 


EQUATIONS  or  THE  SECOND  DEGREE.       321 

previous  to  this  time  to  a  guard  of  soldiers.  The  interest  upon  this 
sum  for  21  months,  amounted  to  $19G0,  and  the  square  of  the  interest 
for  6  months  exceeded  the  square  of  100'"  part  of  the  sum  by  $288,000. 
Required  the  sum  appropriated  annually  to  the  military  academies,  and 
the  rate  of  interest.  Ans.  816,000,  sum  ;   7  per  cent,  interest. 

47.  A  man  in  Cincinnati  purchased  10,000  pounds  of  bad  pork,  at 
1  cent  per  pound,  and  paid  so  much  per  pound  to  put  it  through  a 
chemical  process,  by  which  it  would  appear  sound,  and  then  sold  it  at 
an  advanced  price,  clearing  8450  by  the  fraud.  The  price  at  which  he 
sold  the  pork  per  pound,  multiplied  by  the  cost  per  pound  of  the 
chemical  process,  was  3  cents.  Required  the  price  at  which  he  sold  it, 
and  the  cost  of  the  chemical  process. 

A71S.  He  sold  it  at  G  cents  per  pound,  and  the  cost  of  the  process 
was  i  cent  per  pound. 

48.  The  fore  wheel  of  a  wagon  makes  12  more  revolutions  than  the 
hind  wheel,  in  going  240  yards;  but  if  the  circumference  of  each 
wheel  be  increased  one  yard,  the  fore  wheel  will  make  only  8  more 
revolutions  than  the  hind  wheel,  in  the  same  space.  Required  the 
circumference  of  each. 

Ans.  Circumference  of  fore  wheel,  4  yards ;  circumference  of  hind 
wheel,  5  yards. 

49.  In  the  year  1853  there  were  a  certain  number  of  Woman's 
Rights  conventions  held  in  the  State  of  New  York.  If  6  be  added  to 
the  number  and  the  square  root  of  the  sum  be  taken,  the  result  will  bo 
exactly  equal  to  the  number.     Required  the  number.  Ans.  3. 

How  does  the  negative  solution  arise  ?     Why  is  it  neglected  ? 

50.  A  planter  purchased  a  number  of  slaves  for  $36,000.  If  he  had 
purchased  20  more  for  the  same  sum,  the  average  cost  would  have  been 
$150  less.     Required  the  number  of  slaves,  and  their  average  price. 

Ans.  60  slaves ;  average  price,  $600. 

51.  A  planter  purchased  a  number  of  slaves  for  m  dollars.  If  he 
had  received  n  more  for  the  same  sum,  their  average  price  would  have 
been  c  dollars  less.  Required  the  number  of  slaves  and  their  average 
price. 

V 


822  GENERAL  PROBLEMS  INVOLVING 


n  /4:nm  +  en'      ,,  n  / inm -\- cn^ 

2     '    V  4c        '  2         V  4c       ' 

x'  and  a;"  exjiressing  the  uimiber  of  slaves.     Then,  —,  and  -^  will  ex- 
press the  average  price. 

Now,  when  n  is  zero,  the  two  values  of  x  ought  both  to  be  infinite, 
indicating  an  absurdity.  But  the  expressions  will  not  point  out  an 
absurdity   unless   reduced   to   their   lowest   terms   by   extracting   the 

indicated  roots.     Then  the  two  values  of  x  may  be  written  —  —  dz 


m 


—  -| -^,  plus  any  other  terms  containing  the  higher  powers 


en 


( 

of  X  in  the  denominators.) 

Now,  make  n  =  0,  and  the  two  values  of  x  both  become  infinite. 
An  expression  can  never  be  correctly  interpreted  unless  it  is  reduced 
to  its  lowest  terms,  for,  as  in  the  present  instance,  there  may  be  a 
common  factor  in  all  its  terms,  and  a  particular  hypothesis  made  upon 
that  common  factor  may  lead  to  absurd  results. 

What  do  the  values  become  when  c  =  0  ?  what  when  n  :i=  0  and 
c  =  0  ? 

52.  The  field  of  battle  at  Buena  Vista  is  6  J  miles  from  Saltillo. 
Two  Indiana  volunteers  ran  away  from  the  field  of  battle  at  the  same 
time ;  one  ran  half  a  mile  per  hour  faster  than  the  other,  and  reached 
Saltillo  5  minutes  and  54^*^  seconds  sooner  than  the  other.  Kequired 
their  respective  rates  of  travel.  Ans.  6,  and  5^  miles  per  hour. 

53.  The  New  York  shilling  is  12|  cents.  A  merchant  bought  a 
quantity  of  cloth  for  $60.  The  number  of  shillings  which  he  paid 
per  yard  was  to  the  number  of  yards  he  bought  as  1  to  -i-ff^.  llcquired 
the  number  of  yards  and  the  price  per  yard. 

Ans.  48  yards ;  10  shillings  per  yard. 

54.  A  grocer  sold  1000  pounds  of  coffee  and  1500  pounds  of  rice 
for  $240 ;  but  he  sold  500  pounds  more  of  rice  for  $80  than  he  sold 
of  coffee  for  $60.     Required  the  price  per  pound  of  the  cofiee  and  rice. 

Ans.  Cofi"ee,  12  cents  per  pound ;  rice,  8  cents  per  pound. 

55.  A,  B,  and  C,  entered  into  partnership,  and  gained  as  much  as 
their  joint  fund,  wanting  $200.     A's  gain  was  $240.     He  invested 


EQUATIONS  OP  THE  SECOND  DEGREE.       323 

$100  more  than  B ;  and  the  joint  investment  of  B  and  C  was  S700. 
Required  the  gain  per  cent.,  and  the  amount  invested  by  each 
partner. 

Ans.  80  per  cent.     A's  capital,  S800 ;  B's,  6200 ;  C's  3500. 

56.  There  is  a  vessel  containing  25  gallons  of  -wine ;  a  certain  quan- 
tity is  drawn  out,  and  its  place  supplied  by  water.  As  much  is  drawn 
out  of  the  adulterated  liquor  as  was  first  drawn  out  of  the  pure  wine, 
and  there  is  now  but  16  gallons  of  pure  wine  left  in  the  vessel.  Re- 
quired the  quantity  of  pure  wine  drawn  out  each  time. 

Ans.  First  draught,  5  gallons ;  second  draught,  4  gallons. 

57.  Same  problem  as  last,  except  that  the  vessel  contains  m  gallons 
of  wine,  and  that,  after  four  draughts,  —  gallons  of  pure  wine  are  left. 

.     Ans.  Inrst  draught,  m — \/m;  second, = — ;  third, ; 

_  ^/m  m 

.    _^,     m — y/m 
fourth,  ■ 

These  results  can  readily  be  verified ;  for,  when  the  four  draughts 
are  taken  from  m,  there  will  remain 

—       ^m  —  m        ^m  —  m        ^m  —  m      ,.  ,    .  ,  

\/m  -I = —  H 1 ^^— —   which  is  equal  to  v/w 

+  i  —  vm  -\ —  —  1  -f =^     or     ^=^  -f — ,      or 


_  y/in  ^m^  ^m 

v^m  -|-  1  — v^wi         1 
,  or  — . 

m  m 

The  equation  to  be  solved  was  really  one  of  the  fourth  degree ;  but, 
owing  to  its  peculiar  form,  it  was  readily  reduced  to  an  equation  of  the 
second  degree.  Only  one  of  the  four  values  of  the  unknown  quantity 
have  been  used. 

58.  Same  problem  as  last,  except  that  the  vessel  contains  64  gallons 
of  wine,  and  that,  after  four  draughts,  the  gLth  of  a  gallon  is  left. 
Ans.  First  draught,  56  gallons ;  second,  7 ;  third,  || ;  fourth,  ^f^. 

Verify  these  results  by  adding  them  together,  and  subtracting  their 
sum  from  64. 


324  GKNEIIAL    PROBLEMS     INVOLVING. 

59.  Two  cotton  merchants  rent  a  house  for  a  certain  sum,  with  the 
understanding  that  each  shall  pay  in  proportion  to  the  number  of  bales 
of  cotton  he  puts  in  the  house.  A  puts  in  500  bales-,  and  B  as  many 
bales  as  makes  his  proportion  of  pay  amount  to  $200.  B  afterwards 
puts  in  500  more  bales,  and  then  his  proportion  amounts  to  $225.  Re- 
quired the  sum  paid  for  house  rent,  and  the  number  of  bales  first  put 
in  by  B.  Ans.  House  rent,  $300.     B  first  put  in  1000  bales. 

60.  A  gentleman  has  two  sums  of  money  at  interest,  amounting  to 
$1150.  The  larger  sum,  being  put  out  stii  per  gent,  less  advanta- 
geously than  the  smaller,  brings  only  the  same  amount  of  yearly  inte- 
rest. At  the  end  of  ten  years,  the  smaller  sum,  added  to  its  simple 
interest  for  the  whole  period,  is  to  the  larger  sum,  added  also  to  its 
simple  interest,  as  11  is  to  11 2 .  Required  the  two  sums,  and  the  per 
cent,  on  each. 

Ans.  $550  at  10  per  cent.,  and  $600  at  9i  per  cent. 

61.  A  gentleman  bought  a  rectangular  piece  of  laud,  giving  $10  for 
every  yard  in  its  perimeter.  If  the  same  quantity  of  ground  had  been 
in  a  square  shape,  it  would  have  cost  him  $20  less.  And,  if  he  had 
bought  a  square  piece  of  land,  of  the  same  perimeter  as  the  rectan- 
gle, it  would  have  contained  6|^  square  yards  more.  Required  the 
sides  of  the  rectangle.  Ans.  4  and  9  yards. 

G2.  A  and  B,  together,  invested  $800  in  a  speculation.  A's  money 
was  employed  3  months,  and  B's  5  months.  When  they  came  to  set- 
tle, A's  capital  and  profits  amounted  to  $451 ;  B's  amounted  to  $375. 
Required  the  capital  of  each,  aud  their  gain  per  cent, 

Ans.  A's,  $440 ;  B's,  $360.     10  per  cent.  gain. 

63.  The  seventh  page  of  a  treatise  on  Analytical  Geometry  has  11 
more  lines  than  the  twentieth  page  of  a  treatise  on  Optics ;  but  the 
seventh  page  of  the  Analytical  Greometry  has  4  letters  less  in  each 
line  than  there  are  lines  in  the  twentieth  page  of  the  Optics,  whilst  the 
twentieth  page  of  the  Optics  has  as  many  letters  in  each  line  as  there 
are  lines  in  the  seventh  page  of  the  Optics.  The  total  number  of  let- 
ters on  both  pages  is  3542.  Required  the  number  of  lines  and  letters 
in  each  page. 

Ans.  Analytical  Geometry,  46  lines,  and  42  letters  in  each  line; 
Optics,  35  lines,  and  46  letters  in  each  line. 


EQUATIONS    OF    THE    SECOND    DEGREE.  325 

64.  Required  to  divide  the  quantity,  a,  into  two  such  parts,  that  the 
sum  of  the  quotients,  arising  from  dividing  each  part  by  the  other, 
shall  be  equal  to  m. 

.        T^.    ^        ..    ^    ,    <^      /"«^  —  ^          f         (i      /'"  —  2 
Ans.  First  part,  — -  +  --\  /  ,  or \  /  —  : 

^     '   2  ^  2  V  "i  +  2'         2         2  V  w(  +  2  ' 

,             a         a       I  )ti  —  2          a         a       /m  —  2 
second  part,  —  —  -r-\  / -,  or  — -  +  — -\  / . 

^     '2         2  V  «^  +  2'         2    '    2  V  7n  +  2 

Are  there  four  independent  values?  When  only  will  these  values 
be  real  ?  When  will  one  always  be  negative  ?  When  will  the  two 
parts  be  equal  ? 

65.  Required  to  divide  10  into  two  such  parts,  that  the  sum  of  the 
quotients,  arising  from  dividing  each  part  by  the  other,  shall  be  equal 
to  2Jffi.  Ans.  7  and  3,  or  3  and  7. 

66.  Required  to  divide  the  number  100  into  two  such  parts,  that  the 
sum  of  the  quotients,  arising  from  dividing  each  part  by  the  other, 
shall  be  equal  to  2.  ^Ih.s.  50  and  50. 

67.  Required  to  divide  15  into  two  such  parts,  that  the  sum  of  the 
quotients,  arising  from  dividing  each  part  by  the  other,  shall  be  equal 
to9f  .4h.s-.  13 i,  and  li. 

68.  Four  numbers  are  in  a  continued  proportion,  each  number  being 
an  exact  number  of  times  greater  than  that  which  precedes.  The  dif- 
ference between  the  means  is  8,  and  the  difference  between  the  ex- 
tremes 28.  Required  the  sum  of  the  means,  and  the  terms  of  the 
proportion. 

Ans.  Sum  of  the  means,  24.     The  numbers  are  4,  8,  16  and  32. 
In  this  example,  let  the  unknown  quantity  be  the  sum  of  the  means. 

69.  Same  example  as  preceding,  except  making  the  difference  of  the 
means,  a,  and  the  difference  of  the  extremes,  h. 

Ans.  Sum  of  the  means  +  o  \  /^ -r--.  greater  mean  a  + 

V   6  —  6a     ^ 


/  h  -\-a             -,                 /  h  -\-  a  h 

a  K/  ^ ;    smaller,    a  W     a;  greater   extreme   -^  -f 


Ma^  +  b^(b  —  3a)  „  b         ,        /4a'  -f  b' 

V riTs^, ^;  smaller-- +1\/-^^ 


28 


\ 


326       EQUATIONS  OF  THE  SECOND  DEGREE. 

The  negative  values  liave  been  rejected. 

When  will  these  values  become  infinite?     When  imaginary  ? 

70.  There  are  two  numbers,  such  that  the  square  of  the  first  added 

to  their  product  is  equal  to  m,  and  the  square  of  the  second  added  to 

their  product  is  equal  to  n.     What  are  the  numbers  ? 

,        „.  m  ^     ,  n 

A71S.  First,  =b  —  _  ;  second, 


^m  -\-  n  ^m  -\-  n 

71.  Two  numbers  are  to  each  other  as  m  to  n,  and  the  square  of  the 
first  added  to  the  product  of  its  first  power  by  m,  is  equal  to  the  square 
of  the  second,  added  to  the  product  arising  from  multiplying  its  first 
power  by  n.     What  are  the  numbers? 

Ans.  First,  0,  or  —  m;  second,  0,  or  — n. 

72.  The  sum  of  the  squares  of  two  numbers  diminished  by  twice 
their  product  and  by  twice  the  first  number,  is  equal  to  unity;  and  the 
sum  of  their  squares  added  to  the  first  power  of  the  second  number,  is 
equal  to  twice  the  product  of  the  numbers.     What  are  the  numbers  ? 

Ans.  First,  0,  or  —  | ;  second,  —  1,  or  —  -^. 


TRINOMIAL  EQUATIONS. 

353.  A  trinomial  equation  is  one  of  the  form,  x^"  +  x"  =  q,  involving 
but  one  unknown  quantity  and  three  terms,  and  having  the  unknown 
quantity  in  one  term  affected  with  an  exponent  double  of  that  with 
which  it  is  affected  in  the  other.  A  trinomial  equation  then  contains 
two  terms,  in  which  the  unknown  quantity  enters,  and  an  absolute 
term. 

Let  it  be  required  to  solve  the  equation,  x'^  +  cc^  =:  20. 

Let  x^  =  y.     Then  the  equation  becomes  y^  -\-  y  =  20. 

The  last  equation  gives  y  =  +  4  and  y  =  — 5.  But,  x^  =  y. 
Hence,  x"  =  4,  or  —  5.  Then,  x'  =  +  2,  x"  =  —  2,  x'"  =  +  V  5^ 
x""  =  —  ^/ — 5.  Either  of  the  four  values  will  satisfy  the  given 
equation,  a-''  -\-  x^  =  20.  The  substitution  of  the  last  value  gives 
(—  x/— 5/  +  (—  ^^^bf  =  20,  or,  25  —  5  =  20,  a  true  equation. 

Solve  the  equation,  x^  —  x^  =  702.  Make  x^  —  y.  Then  the  equa- 
tion becomes  y"^  —  ?/  =  702.  From  which,  ?/=-{-  27,  or,  —  26.  But 
x^  =  y.     Hence,  x  =  4/27  =  3,  and  x  =  ^ —  26. 

There  are  really  four  more  values  for  x,  since,  as  will  be  shown  here- 
after, the  number  of  values  is  exactly  equal  to  the  degree  of  the  equa- 


EQUATIONS    or    THE    SECOND    DEGREE.  327 

tion.  But  the  method  of  determining  the  other  values  belongs  pro- 
perly to  the  general  theory  of  equations.  The  two  values  found  can 
be  readily  verified. 

The  examples  given  are  of  a  simple  character.  The  equations  were 
already  of  the  proposed  form.  But  it  frequently  happens  that  an  arti- 
fice must  be  employed  to  put  the  equation  under  the  form  of  x^"  -|-  a;" 
=  q.  Take,  as  an  example,  x^  -f  ^x^  4-  9  =  21.  Add  9  to  both 
members,  and  we  have  x^  -\-^  +  ^x^  +  9  =  30.  Make  a:^  +  9  =  /, 
then,  v^a;^  -f  9  =  y,  and  the  equation  becomes  y^  ■\-  y  =  30.  From 
which  we  get,  ?/  -^  -f  5,  or  —  6.  Then,  a;^  +  9  =  25,  or,  j;^  -f  9  =  36. 
The  values  of  x  are  -f  4,  — 4,  +  v/27,  and  —  v/27.  The  negative 
sign  of  the  radical,  ^az  -(-  9,  must  be  taken  in  connection  with  the  last 
two  values.     This  is  indicated  by  the  equation,  y/x^  -f-  9  =  y  =  —  6. 

Again,  take  ^^^^'   +  1  =  '^-^^  +  01- 
This  may  be  written,  ^:i±^'  —  (^!±J^'  =  90. 

XX 

a-^4-  9 
Let =r  ?/.     The  equation  then  becomes  if  —  ?/  =  90.     Hence, 

2/  =  -f-  10,  or  —  9.     Then,  ^-^  =:.  10,  and  "^^  =  —  9.      These 

9  .9 

equations  gives  the  four  values,  -f  9,  -f  1,  and  —  -   -\-  2  ■</45,  —  ~ 

— i  V'45: 


The  equation,  i/x  -f  1  -f  X/x^  -1-  2x  +  1  =  20,  ,may  be  changed 
into  ^x  -\-l+  Vx  -f-  1  =  20.  Let  Va;  +  1  =  y.  Then  ^/x  +  1 
=  2/^  and  the  equation  becomes  %f  -{■  y  =  20.  From  which,  ?/  =  4,  or, 
—  5.  Then  ^/x  +  l  =  16,  and  ^x  -f  1  =  25.  Hence,  x  =  255, 
or,  x  =  624. 

The  equation,  x^  -f  ^x^  -f  56  =  34,  may  be  placed  under  the  pro- 
posed form  by  adding  56  to  both  members.  Then,  x^  -j-5Q-{-\/x^  -f-  56 
=  90.  Let  Vx^  +  56  =  y.  The  equation  then  becomes  y^  +  y  =  90. 
From  which,  y  =  -f  9,  or,  —  10.  Then  a;  =  -f  5,  —  5^  4.  y  4T, 
--^/44^ 

These  illustrations  are  sufiicient  to  explain  the  spirit  of  the  process. 
No  general  rule  can  be  given.  Any  modification  may  be  made  upon 
the  equation  that  will  place  it  under  the  proposed  form. 


328  EQUATIONS     OF    THE    SECOND    DEGREE. 

GENERAL   EXAMPLES. 

1.  Solve  the  equatiou,  a:|  +  xi  =  12. 

Ans.  03  =  +  27,  or  —  64. 

2.  Solve  the  equation,  x  +  xl  =  Cy.  Ans.  a;  =  4,  or  9. 

3.  Solve  the  three  equations, 

2'  +  zy  +  y^  =  1900  (A). 
x''  +  xz  +  z-  =  1300  (B). 
I/'-^^I/+^'=    700(C). 

Subtracting  (B)  from  (A),  and  (C)  from  (B),  and  factoring,  we  get, 

GOO     ,^^        ,  600    ,^^ 

y+z  +  x  =  ■ CD),  nudv  +  z  +  x  = (E). 

1/  —  X  z—y  - 

x  -4-  z 
By  equating  (D)  and  (E),  wc  get  y  —  ^—^ — .     This  value  of »/,  sub- 
stituted in  (D),  gives  z  +  x  =  ■ ,  or,  z~  —  x-^= 800 ;  from  which  z^  = 


x"  +  800.     Substituting  for  z  its  value  in  B,  we  get  ce"  +  x  V  x' +  800 
+  a;2  +  800  =  1300.     From  which,   x  ^x^  +  800  =  500  —  2x\ 
Squaring  and  reducing,  there  results,  S.^"  —  2800x' =  — 250000,  or, 
.       2800x2  250000 


■-^3  3 

The  combination  has  led  to  a  trinomial  equation,  which,  when  solved, 
will  give,  for  one  system  of  values,  x  =  10,  y  =  20,  and  z  =  30. 

4.  Solve  the  equations, 

f  +  ^^  +  x^  =  7, 

if  —  xy  —  a;^  =  —  5. 

Ans.  X  =:  -f  2, —  3,  y  =  -\-\  —  1. 
6.  Solve  the  equations, 

X  +  y  +2  =6, 

x^-\-f-\-^^  =  14, 
xz  =  3. 

Ans.  y  ^  2,  X  =  1,  z  =  S. 
One  system  of  values  only  given. 

6.   Solve  the  equations, 

i/  +  x'=  2600, 

f  —  2y=:2x. 

Ans.  y  =  10,  and  x  =  40. 
One  set  of  values  only  given. 


EQUATIONS    OF    THE    SECOND    DEGREE.  329 

7.   Solve  the  equations,  i/^  +  x^  =  ^  +  2/*  +  B, 

A71S.  One  set  of  values,  y  =  2,  and  x  =  6. 


8.  Solve  the  equation,  ^x^  +  4+  .^a;-  +  4  =  6. 

Ans.  One  set  of  values,  x'  =  +  \/12,  x"  =  —  ■v/12 


9.  Solve  the  equation,  y/y'—y  +  V/ — y  =3v/10  +  </90. 
^l??s.  One  set  of  values,  y  =  10,  or  y  =  —  9. 
The  positive  value  of  the  radical  only  is  taken. 

10.  Solve  the  equation,  x^  +  7  +  y/ x^  +  7  =  20. 

Ans.   x'  =  +  3,   x"  =  —  3,  ./;'"  =  +  v/"T8,  .c'-  =  —  v/lS: 

PROBLEM  OF  THE  LIGHTS. 

354.  Two  lights  are  placed  on  the  same  indefinite  right  line,  the  one 
shining  with  an  intensity  represented  by  a,  the  other  with  an  intensity 
represented  by  b ;  the  problem  is  to  find,  on  this  indefinite  line,  the 
point  or  points  of  equal  illumination  —  assuming  a  principle  of  optics, 
that  the  intensity  of  a  light  varies  inversely  with  the  square  of  the 
distance  from  that  light. 

A  IB 


Let  A  be  the  position  of  the  first  light ;  B,  that  of  the  second. 

Let  m  =  AB,  the  distance  between  the  lights. 

Let  a  =  intensity  of  first  light,  at  one  foot  from  A. 

Let  b  =  intensity  of  second  light  at  one  foot  from  B. 

Let  I  be  the  unknown  point  of  equal  illumination. 

Let  B  I  =  ??(-  —  x  =  distance  of  same  point  from  B. 

Now,  in  accordance  with  the  assumed  optical  principle,  the  intensity 
of  the  first  light,  two  feet  from  A,  will  be  expressed  by  -7-\for  2^  :  l'^  :  : 
a  :  -r),  at  three  feet,  by  -r-,  and  at  x  feet,  by  —^,  and  this  last  expres- 
sion will  represent  the  intensity  of  the  first  light  at  the  unknown  point, 
I.  In  like  manner,  the  intensity  of  the  second  light,  at  B,  will  be 
28  * 


330  EQUATIONS    OF    THE    SECOND    DEGREE. 

denoted  by  r^.     But,  by  the  conditions  of  the  problem,  the  in- 

(vM  —  xy  ''  ^         ^ 

tensities  of  the  two  lights  must  be  equal  at  I.     Hence,  wc  have  —^  = 

r„,  which  may  be  chauo-cd  into —  =  — ,  and,  by  extract- 

0"  —  xj  -^  °  (?)i  —  a';)2        6 '        ^   -> 

ii)<;-  the  root,  we  get =  rb  — -.    We  will,  for  convenience,  call 

m  —  x  ^i 

.r'  that  value  of  x  which  is  connected  with  the  positive  sign  in  the 
second  member,  and  we  will  call  od'  that  which  is  connected  with  the 

negative  sign.     Hence, -.  =  — ^,  and r,  =  —  — ^.    And 

m  —  x        ^i  m  —  x"  ^5 

1  •      xi  1-  ,     ,  m^/a  ,    „  my/ a 

solving  these  equations,  we  get  x:  =     _    — —^  and  x''  =  — =:--^^ — -^. 

Now,  since  there  are  two  distinct  values  for  rr,  we  conclude  that  in 
general  there  will  be  two  points  of  equal  illumination.  By  this  we  do 
not  mean  that  there  will  be  two  points  of  equal  brilliancy,  but  that  there 
will  be  two  points  where  the  intensities  of  the  two  lights  will  be  equal 
to  each  other.     By  subtracting  cc'  and  x"  in  succession  from  m,  we  get 

,  mh  .  ,,  m^h 

^       -'  -  and  —        -" 


y/a  +  y/h  ^a  —  y/h 

Hence,  we  have  the  system  of  values: 

x'  =  — ^= :=,  distance  from  A  to  first  point  of  equal  illumination. 

v/a  +  v/6 

,  m^h  T  «  -r. 

m  —  x  =  — = m,  distance  from  B  to  same  point. 

^/a+  ^h 

x"  =  —^ -=,  distance  from  A  to  second  point  of  equal  illumina- 

m  —  x"  = := -^_,  distance  from  B  to  same  point. 

Now,  these  expressions  have  been  deduced  upon  the  supposition  that 
the  point,  I,  was  between  A  and  B,  and,  consequently,  we  have  assumed 
that  the  distance  A  I  is  positive,  when  estimated  on  the  right  of  A, 
and  the  distance  B  I  positive,  when  estimated  on  the  left  of  B.  Hence, 
if  either  x'  or  ic"  becomes  negative  in  consequence  of  any  imposed 


EQUATIONS  OF  THE  SECOND  DEGREE.       331 

condition,  the  corresponding  point  of  equal  illumination  will  be  found 

on  the  left  of  A.      And,  in  like  manner,  if  either  m  —  x'  or  m  —  x" 

becomes  negative,  the  point  will  be  found  on  the  right  of  B. 

We  will  begin  the  discussion  by  supposing  a  =  b.     Then,  x'  = 

mVa  ms/a        m  AB  •   x  •     i,  u- 

— = =.  =  =  =  -C-,  or  A  i  =  — -— .     Hence,  the  point  is  halt 

Va-\-Vb        2v/a         ^  ^ 

way  between  the  lights,  as  it  obviously  ought  to  be.     Now,  m  —  x', 

which  expresses  the  distance  from  B  to  I,  ought  to  give  the  same  point. 

,    ,  „  ,  «i-«/6  niy/b        m       ^^        ... 

It  does  so,  for  m  —  x  =  -rm =r-  = =  =  -r--       >v  e   will   next 

Va+  ^b       2V6         ^ 

examine  whether  there  can  be  a  second  equally  illuminated  point,  wheu 
the  intensities  of  the  two  lights  are  the  same.     Wc  have  x"  = 

,  „  vi\/b  ^,  ,         .    ,.  , 

=  cc,  and  m  —  x    = jr—  = —  oo.     ihcse  values  indicate  that 

the  second  point  is  at  an  infinite  distance  on  the  right  of  B ;  or,  in 
other  words,  that  there  is  no  such  point  at  all.  This  ought  to  be  so, 
for,  when  the  two  lights  are  of  equal  intensity,  the  one  cannot  throw  its 
beams  with  the  same  power  to  a  greater  distance  than  the  other,  and 
this  must  be  the  case  if  there  be  a  second  point.  The  second  values 
(x"  and  m  —  x")  then  denote  impossibility.  By  going  back  to  the 
equation  of  the  problem,  wc  see  that  it  will  become,  when  a  =  b,  x^  = 

(jn  —  xy,  an  equation  which  can  only  be  true  when  m  =  0,  or  x  =  — . 

But  wi  =  0  is  contrary  to  the  hypothesis ;  hence,  x  =  -—  is  the  only 

true  value.  The  assumption  then,  that  there  was  a  second  point  (a 
being  equal  to  b),  was  absurd,  and  led  to  oo ,  the  appropriate  symbol  of 
absurdity. 

"We  will  nest  suppose  a  >  i. 

Then,  x'  =  —= n  becomes  ">  — ,  because,  if  the  denominator 

^/a  +  v/i  2 

m 

were  2  ^/a,  the  value  of  the  fraction  would  be  "i^,  and  since  the  deno- 

VI 

minator  is  less  than  2^  a,  the  value  is  greater  than  tt- 

Hence,    the    point,   I,   is    nearer    B    than    A.       In    this    case, 
A  IB  r 


SoZ  EQUATIONS  OF  THE  SECOND  DEGREE. 

m  —  x'  <^  ^,  for,  were  the  denominator  of  the  fraction  — ~— ^ —  __ 

m 

exactly  'Z\/b,  the  value  of  the  fraction  would  be  ^^J  t)ut  since  the  deno- 
minator is  greater  than  2,  ^h,  the  value  is  less  than  — .       This    result 

corresponds  to  the  former,  and  places  I  nearer  B  than  A ;  that  is, 
nearer  the  light  of  feebler  intensity,  as  it  ought  to  do.  Wc  see  that  x" 
is  greater  than  m,  for  were  the  denominator  of  x",  the  -s/a,  its  value 
would  be  equal  to  m,  but  as  y/  a  —  y/  b  is  less  than  V  a,  x"  is  greater 
than  m.     Hence,  the  second  point  of  equal  illumination  is  beyond  B. 

T      1  .  ,f  m  y/  b        ^  .  .  , 

In  this  case,  m  —  x    = = =^   becomes    negative,  since   the 

^a—^/b  ^ 

numerator  is  negative  and  denominator  positive,  and,  therefore,  the 

second  point  must  be  on  the  right  of  B.     The  results  then  agree  with 

each  other,  and  agree  with  the  fact.      For,  after  passing  the  point  I,  the 

intensity  of  the  1st  light  becomes  feebler  than  that  of  the  2d.     At  the 

point  B,  there  is   the    greatest  diflference    between   their  intensities. 

Beyond  B,  both  lights  diminish  in  brilliancy,  but  the  2d  more  rapidly 

than  the  1st,  because  of  its  greater  feebleness;  and  we  at  length  reach  a 

point,  V,  where  the  illumination  is  equal. 

Next,  suppose  h^  a. 

Then,  x!  r=  — ^;= ^  <r  — .     For,  were  the  denominator  2  ^TT,  the 

value  of  the  fraction  would  be  ~,  but  as  the  denominator  is  ^  2  y/~a, 
the  value  of  the  fraction  would  be  <^  — 


I" 


Hence,  the  point,  I,  will  be  found  nearer  A  than  B.  We  have  m  —  x'  = 
^  ~2  ;  for,  were  the  denominator  2  ^Ij^the  value  would 


^/a  +  ^b 
m 


be  =  -  ,  but  as  the  denominator  is  <^  2  ^b,  the  value  is  ^  — .   The 

Z  2 

results  then  agree,  and  place  the  point,  I,  nearer  the  feebler  light. 

Again,  we  have  x"  =    — = =.  affected  with  the   sign  minus,  the 

■v/a  —  ^b 


EQUATIONS    OF    THE    SECOND    DEGREE.  333 

numerator  being  positive,  and  denominator  negative.  The  second  point. 
I",  must  then  be  on  the  left  of  A.  By  multiplying  the  numerator  and 
denominator,  of  the  value  of  m  —  x",  by  minus  unity,  we  will  have  ??i  — 

x"  =  — =^ ^  m  ;  for,  were  the  denominator  z=z  y/  b,  the  value 

■s/h  —  y/a 

would  be  equal  to  m,  and  since  the  denominator  is  <^\/b,  the  value  is 
greater  than  m.  Hence,  the  second  point  is  beyond  A  at  1",  as  has 
just  been  shown.  Were  the  values  of  m,  a,  and  h,  given,  the  position 
of  the  point  of  equal  illumination  could  be  readily  determined.  Let 
m  =  20  feet,  a  =  16,  6  =  36.  Then  x' =  ^  feet,  and  m  —  x'  =  12 
feet,  and  the  intensity  of  the  two  lights  will  be  equal  at  the  point  cor- 
responding to  these   distances.     For,  the  intensity  of  the  1st  will  be 

^,  =  \,  and  that  of  the  2d  ^^  =  \ 

I"  A  I  B 


It  can  readily  be  shown  that,  between  I  and  A,  the  intensity  of  the 
1st  light  is  greater  than  that  of  the  2d.  Thus,  at  2  feet  from  A,  that 
of  the  1st  light  will  be  expressed  by  4,  and  that  of  the  2d  by  -^.  At 
the  point  I",  40  feet  on  the  left  of  A,  and  60  feet  from  B,  the  intensi- 
ties of  both  lights  will  be  expressed  by  jjjjth,  and  arc,  therefore,  equal. 
And,  by  assuming  other  points,  and  calculating  the  corresponding 
intensities,  we  would  find  that  they  were  only  equal  at  I  and  I". 

Now,  let  m  =  0,  and  a  unequal  to  h,  then  x',  m  —  x',  x",  and 

m  —  x"  =  0.     These  are  plainly  absurd  solutions,  for  the  two  lights, 

shining  with  unequal  brilliancy,  cannot,  equally,   illumine   the  point 

at  which  both  are  placed.    By  recurring  to  the  equation  of  the  problem, 

x^        a 
it  becomes,  when  m  =0,  — ,  =  — .       This  equation  can  only  be  true 

when  a=zh,  which  is  contrary  to  hypothesis. 
Now,  let  m  =  0,  and  a  =  h. 

Then,  x'  =  0,  m  —  x'  =  0 ;  x"  =  —,  and  m  —  x"  =  — .     The  first 

two  values  (x'  and  m  —  a;',)  show  that  the  point  at  which  the  lights  are 
placed,  is  equally  illuminated.  The  second  two  (x"  and  m  —  x",)  are 
the  symbols  of  indetermination.  This  ought  to  be  so,  for  when  the 
equal  lights  are  placed  together,  the  points,  at  one  foot,  two  feet,  three 
feet,  &c.,  are  as  much  irradiated  by  one  light  as  the  other.     There  is, 


334  EQUATIONS    OF    THE    SECOND    DEGREE. 

therefore,  no  particular  determinate  point  at  wliich  the  intensities  are 
equal,  and  we  say  that  the  problem  is  indeterminate.     By  recurring  to 

the  ef|uatiou  of  the  problem, =  -—,  we  see  that  when  m  =  0, 

^  ^  (in  —  xy        b 

and  a=h,it  reduces  to  an  identical  equal  equation,  x^  =  x^,  which  can 

be  satisfied  for  any  values  of  x. 

Whenever  we  get  tt  as  a  solution  to  a  problem,  we  can  tell,  by  recur- 
ring to  the  original  equation,  whether  it  indicates  indetermination.  If 
the  particular  hypothesis,  which  reduces  the  solution  to  — ,  also  reduces 

the  original  equation  to  an  identical  form,  we  have  a  case  of  indeter- 
mination. But,  if  the  given  equation  do  not  reduce  to  an  identity,  we 
have  a  vanishing  fraction. 

Another  test,  also,  may  be  employed.  If  the  value  becomes  -r-  in  con- 
sequence of  a  single  hypothesis,  we  know,  certainly,  that  we  have  a 
vanishing  fraction,  and  not  an  indeterminate  solution.     The  preceding 

values  become  -^  in  consequence  of  the  two  hypotheses,  m  =  0,  and 
a  =  h,  and  might  or  might  not  be  vanishing  fractions.  Thus,  the  ex- 
pression, \ r^--rhi — To^,  is  a  double  vanishing  fraction  for  a;  =  a,  and 

^  {x  —  a)  (x^ — Iry 

X  =:h. 

Now,  suppose  the  first  light  extinguished ;  then,  a  =  0 ;  and  we 
find  x'  and  x"  =  0,  and  m  —  x,  and  on  —  x"  =  m.  These  solutions, 
at  first,  seem  absurd,  since  they  indicate  that  the  point  of  equal  illumi- 


nation is  at  A.  But,  suppose  the  light  at  A  to  be  very  feeble,  there 
will  then  be  two  points  of  equal  illumination,  I  and  F,  very  near  to 
and  on  opposite  sides  of  A.  Make  the  first  light  still  more  feeble,  and 
the  two  points,  I  and  I',  will  approach  nearer  to  A ;  and,  finally,  when 
that  light  is  extinguished,  they  will  unite  at  A. 

Next,   suppose  both    lights  extinguished,   or  a  =  0,   and  b  =  0. 

Then,  x',  x",  m  —  x',  and  m  —  x"  become  — -,  indeterminate,  as  they 

ought  to  be. 

Suppose  m  =  cc,  or  that  the  lights  are  infinitely  distant  from  each 


UNDETERMINED    COEFFICIENTS.  335 

other.  Then,  if  a  and  h  are  finite,  the  values  all  become  infinite,  as  they 
ought,  since  there  can  be  no  point  of  equal  illumination.  The  equa- 
tion  of  the  problem,  —.  = ,  can  be  satisfied,  when  m  ==  oc, 

^  x^       (m  —  xy 

by  making  x  =  0,  which  places  the  point  of  equal  illumination  at  A ; 
or,  by  making  x  =  m,  which  places  the  point  of  equal  illumination 
atB. 

Suppose,  a  :=  go;  then,  x'  and  x"  both  become  equal  to  m,  since 
v^6  may  be  neglected  in  the  denominators  of  the  fractions,  and  the 
point  of  equal  illumination  is  then  at  B.  The  values  of  m  —  x'  and 
m  —  x"  become  zero,  and  indicate  the  same  point.  This  ought  to  be 
so ;  for,  by  making  the  intensity  of  one  light  infinitely  great,  wc  make 
that  of  the  other  relatively  infinitely  small,  and  we  then  ought  to  get 
the  same  result  as  we  did  when  one  of  the  lights  was  supposed  to  be 
extinguished. 


UNDETERMINED    COEFFICIENTS. 

355.  The  method  of  undetermined  coefilcients  is  used  to  develop 
algebraic  fractions  into  a  series,  and  to  determine  the  value  of  the  con- 
stants which  enter  into  identical  equations  j  that  is,  equations  which  can 
be  satisfied  by  any  value  whatever,  attributed  to  the  unknown  quantity. 

For  the  development  of  fractions  we  assume  the  form  of  develop- 
ment ;  and  we  are  governed  in  our  assumption  of  this  form  by  our 
knowledge  of  what  it  ought  to  be.     Thus,  if  wc   proceed  to  expand 

— -—  by  the  ordinary  process  of  division,  we  obtain  — ■ —  =  1 

Q/   ~j"  X  ill   — r*  OC  rt 

/y.2  ^-,3  ^^4  rv,5 

H s  —  — ,  +  ^  —  — ,  4-,  &c.      The  quotient  is   an   infinite   series, 

arranged  according  to  the  ascending  powers  of  x,  and  having  x  in  each 
term  affected  with  a  positive  and  an  entire  exponent.  The  first  terra  may 
be  supposed  to  contain  x,  affected  with  a  zero  exponent,  and  the  coeffi- 
cients of  x  are  then  1 1 -. -,  -f  &c. ;   and  it  is  evident  that 

these  coefficients  are  entirely  independent  of  the  particular  values  that 

may  be  attributed  to  x.     They  are  still  1, f-  — , s  +  &c.; 

a         (t         « 


336  UNDETKR  MINED    COEFFICIENTS. 

when  X  =  0,  1,  2,  1000,  anything  whatever.  We  may  remark,  in 
this  connection,  that  the  independence  of  the  coefficients  upon  the 
vakie  of  the  letter  with  which  they  are  associated  is  not  confined  to  the 
expansion  of  fractions,  but  is  true  in  all  developments  whatever.  Thus, 
(a  +  xy-  =  a^  -j-  lax  +  oi? ;  the  coefficients  arc  a^,  2a,  and  1,  and 
these  coefficients  will  not  be  altered  by  any  change  in  the  valu6  of  x. 
We  have  seen,  by  perforhiing  the  division,  that  the  exponents  of  x, 

in  the  development  of  '■ — ,  must  be  positive  and   entire,  and  that 

the  coefficients  must  be  independent  of  x.  If,  then,  we  were  required 
to  assume  the  form  of  development  of  this  fraction,  or  any  like  it,  we 
would  know  that  the  assumed  form  must  fulfil  the  required  conditions. 
Now,  since  a  may  be  regarded  as  the  coefficient  of  a",  the  fraction, 

,  may  be  considered  as  arranged  with  reference  to  x.  both  in  the 

a  -\-x        •' 

numerator  and  denominator;  and  we  have  seen  that  a  fraction  so  ar- 
ranged, beginning  with  the  zero  power  of  x,  gives  positive  and  entire  ex- 
ponents for  X  in  the  quotient.  This  law  can  readily  be  shown  to  be  gene- 
ral, and  it  is  not  even  necessary  that  the  zero  power  of  x  should  enter 
into  the  numerator.  If  the  exponents  of  x  in  the  numerator  are  all  po- 
sitive and  entire,  and  the  terms  arranged  according  to  the  ascending 
powers  of  x,  and  if  the  denominator  is  also  arranged  in  like  manner,  and 
contains  a  constant^  a',  that  is,  a  term  involving  x°,  then  all  the  exponents 
will  be  necessarily  positive  and  entire.     For,  take  the  general  fraction, 

^-"^^f      .  ^.      =  A  -f  'Bx'^  -f  Ca;-""  -{-  &c.,  (N),  in  which  the  second 

member  represents  the  development  of  the  fraction.  Now,  since  a  frac- 
tion and  its  development  must  constitute  an  identical  equation,  it  is 
plain  that,  from  the  nature  of  such  equations,  (N)  must  be  true  when 
£c  =  0.     But  C.x"'',  which  represents  the  term  (if  any)  containing  a 

C 
negative  exponent,  can  be  written  — ,   and  is   equal    to   infinity  when 

X  =  Q.  Hence,  the  whole  second  member  is  infinite.  But,  a;  =  0 ; 
reduce  the  first  member  to  —  ;  n,  on,  p,  and  q  being  supposed  posi- 
tive.    Now,  if  a  ==  0,  we  have  — ^  =  0  =  second  member  =r  oo,  which 

is  absurd.  If  a  be  not  zero,  still  we  have  —  =  oo,  or  a  finite  quantity 
equal  to  an  infinite,  which  is  absurd. 


UNDETERMINED    COEFFICIENTS.  337 

If,  then,  all  the  exponents  in  the  numerator  and  denominator  are 
positive  and  entire,  and  the  denominator  contains  a  constant,  or  both 
numerator  and  denominator  contain  constants,  all  the  exponents  of  the 
development  must  be  positive.  "We  will  now  show  that  all  the  expo- 
nents of  X  must  also  be  entire.  We  have  represented,  by  Bx~,  the 
term,  if  any,  involving  a  fractional  exponent.     Now,  suppose  —  =:  ^. 

Then,  Bx  °  =  Ba;  =  B^/j:.  But  every  square  root  has  two  distinct 
values.  Hence,  ^y/x  has  two  distinct  values,  and  the  whole  second 
member  has  two  distinct  values.  But  no  fraction  containing  only  entire 
exponents  can  give  two  quotients.  The  assumption,  then,  of  a  frac- 
tional exponent  in  the  development  is  absurd.     If  —  =  J,  |,  |,  &c.,  the 

term  Bx  ■>  would  have  three  values  and  three  quotients,  which  is  ab- 
surd.    So,  if  —  =  any  fraction,  there  would  be  more  than  one  quotient. 
m  . 

We  are  now  prepared  to  assume  the  form  of  development  of  any  al- 
gebraic fraction,  arranged  according  to  the  powers  of  a  certain  letter, 
and  having  that  letter  in  the  first  term  of  the  numerator  affected  with 
an  exponent  equal  to,  or  greater  than  that  of  this  letter  in  the  first  or 
last  term  of  the  denominator.  The  exponents  of  the  aiTanged  letter 
are  also  assumed  to  be  positive  and  entire,  both  in  the  numerator  and 
denominator. 

Let  us  then  place  — ; —  =  A  -f-  B.c  +  Gx^  +  J)x^  +  Ex*  +  &c.,  in 
a  -j-  X 

which  A,  B,  C,  D,  &c.,  are  independent  of  x,  and  in  which  all  the  ex- 
ponents of  x  are  positive  and  entire.  A  is  the  representative  of  that 
term  which  does  not  contain  x,  or  contains  x  affected  with  a  zero  expo- 
nent. There  is  generally  such  a  term  in  a  development,  and  there 
must,  of  course,  be  a  representative  of  that  term.  Clearing  of  frac- 
tions and  arranging  according  to  the  ascending  powers  of  x,  we  have 

a  =  Aa  +  a  B  I  i-  -f  rt  C  I  x2  -f  «  D  I  x^  -f  a  E  I  x*-{-&c.  (31). 
+A  I         -f  B  I  +C\  +cl 

Now,  since  the  coefficients  are  supposed  to  be  independent  of  x,  their 
values  will  not  be  affected  by  making  x  —  0.     If  then  we  can  deter- 
mine these  values  when  x  =  0,  they  will  be  true  when  x  =  1,  2,  3 ; 
anything  whatever.     Making  x  =  0,  we  have  a  =  Aa.     Hence,  A  =  1. 
29  w 


338  UNDETERMINED    COEFFICIENTS. 

Now,,  since  a  =  Ka,  tlaese  two  terms  cancel  each  other,  and  (M)  becomes 

+a|      +b|       +c|      +d| 

Dividing  both  members  of  this  equation  by  x,  which  we  have  a  right 
to  do,  we  get 

0  =  a'Q  +  aG  \x  +  aJ)  \x^  -\-  ar&  \x^  +  kc.  (N). 
+A     +B I         +C I  +D I 

Making,  again,  a;  =  0,  wc  have  left,  0  =  aB  -f  A.      Hence,  B  =  — 

A  1 

—  = .     Now,  since  «B  +  A  has  been  found  to  be  zero,  they  may 

a  a 

be  stricken  out  of  (N),  and  that  equation  will  become 

+b|     +c|      +d| 

And  again,  dividing  out  by  x,  we  have 

0=  aQ   +  a  D  I  x^  +  a  E  I  .7j2  +  &c.  (P). 
+B       +C|         +d| 

Making  cc  =  0  in  (P),  we  have  left,  0  =  a  C  +  B.     Hence,  C  = 

B  1 

=  H 2-     Omitting  the  term,  oC  +  B  =  0,  in  equation  (P), 

and  dividing  again  by  x,  we  have 

0  =  a  D  +  a  E  j  a;  +  &c.  (2). 

Making  a:  =  0,  we  have  left,  0  =  aD  +  C.     Hence,  J)  = = 

a 

3 .     Dividing  (2)  by  x,  and  again  making  a;  =  0,  we  have  aE  +  D 

=  0.     Hence,  E  = =  -| — j.     Now,  if  we  substitute  for  A,  B, 

a  a*  '     ' 

0,  &c.,  their  found  values,  in  the  original  equation, =  A  +  Ba- 
ft +  x 

+  Cx^  +  Da-3  +  Ea;^  +  &c.,  we  will  have  — "—^  =  1  _  i^  +  ^  _ 

a   -\-   x  a        a^ 

7?  X^  X^ 

-,  -{ — J r  +  &e. :  the  same  result  that  we   obtained  bv  division 

rr        a*        a"  -^ 

By  transposing  a  to  the  second  member  in  equation  (M),  we  have 
0  =  A  a  I  x°  +  a  B  I  a;  +  a  C  I  a;"  +  rr  D  I  a'''  +  &c. 

— «  +a|        +b|         4-C 


UNDETERMINED    COEFFICIENTS.  ooi> 

And,  since  we  have  found  A«  —  a  =zO,  a  A  4-  A  =  0,  aC  +  B  =  0, 
aD+  C  =  0,  &c.,  we  conclude,  that,  if  ice  have  an  equation  wliose first 
member  is  zero,  and  xohose  second  member  contains  all  its  terms  ar- 
ranged according  to  the  ascending  powers  of  a  certain  letter,  the  expo- 
nents of  this  letter  being  all  positive  and  entire,  and  its  coefficients  in- 
dependent of  it,  these  coefficients  will  be  sep)arately  equal  to  zero. 

This  is  the  first  enunciation  of  the  principle  of  undetermined  co- 
efficients, and  ought  to  be  remembered. 

The  second  enunciation,  (which  is  an  immediate  consequence  of  the 
first),  is  as  follows  :  if  we  have  an  equation  of  the  form,  a-\-hx+  cx^  -|- 
dx^  +  ex*  +  &c.,  =a'  -\-  b'x  +  c'x^  +  ^'^^  +  '^^'^  +  &c.,  which  is  satis- 
fied for  any  value  of  x,  then  the  coefficients  of  the  like  powers  of  x 
in  the  two  members  will  be  respectively  equal  to  each  other. 

For,  by  transposition,  we  have  0  r=  (a'  —  a)  x°  -\-  (Jj  —  b)  x  -\- 
(c'  —  c)  x^  ■\-  (d'  —  d')  x^  -f  (e'  —  c)  x*  -\-  &c.  And,  since  these  coeffi- 
cients are  independent  of  x,  we  must  have,  by  what  has  just  been  shown, 
a'  —  a  =  0,  b'  —  b  =  0,  d  —  c=0,  d'  —  d  =  0,  ^  —  e  =  0,  &e. 
Hence,  a'  =^  a,  b^=b,  cf  =  c,  d  =  d,  e'  =^  e,  as  enunciated. 

The  preceding  equation,  a  +  ix  -f  cx^  -f  dx'  -f  ex*  -f  &c.  =  a'  -f 
b'x  -f  c'x^  -\-  &c.,  and  all  other  equations  to  which  the  second  enuncia- 
tion is  appliablc,  belong  to  a  class  of  equations  called  identical  equa- 
tions, the  two  members  of  which  diffi^r  only  in  form. 

It  is,  in  general,  most  convenient  to  develop  expressions  in  accord- 
ance with  the  second  enunciation. 

1 — X 
Required  the  development  of  . 

Let  J-=^  =  A  -f  B.r  +  Gx'  -{-  J)x'  -{-  Ex^  -f  &e. 

Clearing  of  fractions,  we  have  1  —  x  =  A  -f  (B  +  A)  a;  -f- 
(C  +  B)x'-hCG  +  D)  a;=»  -f  (E  +  D)  X*  -f  &c.     By  placing  a-  =  0,  we 

have  A  =  1,  and  we  might  proceed  as  we  did  with  the  fraction, . 

'  a  -f-  X 

continually  dividing  by  x,  and  making  x  :=  0  in  the  resulting  equation. 

But,  by  the  second  enunciation,  we  have  at  once  A  =  1,B  +  A  =  —  1, 

C  +  B  =  0,  C  +  D  =  0,  and  E  -f  D  r=  0.     From  which,  A  :=  1, 

Br=  —  2,  C  =  +2,  D=  —  2,  E=—  2.     And,  substituting  these 

1 ^ 

values  of  A,  B,  C,  D,  E,  &c.  in  the  equation,  , 1=  A  -f  Bx  +  Cx^ 

1  -\-x 

I X 

+  &c.,  we  get  y-—  =.  1  _  2x  -f  2x^  —  2x^  -f  2x*  —  &c. 


340  UNDETERMINED     COEFFICIENTS. 

This  is  tlic  same  result  that  we  would  obtain  by  actually  performing 
the  indicated  division. 

We  have  assumed  the  form  of  development  to  be  A  +  Bx  +  Cx^  +  &c., 
in  which  the  exponents  are  all  positive  and  entire.  Now,  if  we  have 
an  expression  whose  development  must  necessarily  contain  negative  or 
fractional  exponents,  it  would  be  absurd  to  place  it  equal  to  A  +  B  +  a;  + 
C.T^  +  &c.,  and  the  result  will  make  manifest  the  absurdity  by  the 

symbol,  cc  .     Suppose  it  be  required  to  develop -.     It  is  plain 

that  the  first  term  of  the  development  is  x"' ;  if,  then,  we  attempt  to 
develop  the  expression  by  the  method  of  undetermined  coefficients,  we 
commit  an  absurdity,  and  that  absurdity  ought  to  be  made  manifest  in 
the  result  by  the  appropriate  symbol,  co  . 

Let,  then,  — —,  =  A  +  Bx  +  Cx^  +  J)x^  +  Ex"  +  &c. 
X  —  x/ 

From  this  we  get  l=Ax  +  (B—A)x^-{-(G  —  B)x^-].(D  —  C)x*+&c. 

Now,  in  accordance  with  the  first  enunciation,  make  x  =  0,  and  we 
get  1=0,  an  absurd  result.  But,  if  we  first  divide  both  members  by 
X,  and  then  make  x  =  0,  we  will  have  i  =  go  =  A,  and  the  absurdity 
is  pointed  out  by  its  appropriate  symbol. 

The  above  expression  can  be  developed  by  changing  its  form.  De- 
compose   5  into  its  factors,  —  X  j; ,  then  develop bv 

X  —  a;  X        1  —  X  ^  1  —  X 

the  method  of  undetermined  coefficients,  and  multiply  the  development 
by  — .     By  developing,  we  find      =  1  +  x  +x^  -}•  x^  -\-x*+  &c. ; 

111  ,         „ 

consequently, --^  =  —  X  -, =r  x"'  +  x°  +  x  +  x^  +  ^^  -f 

.x"  +  &c.  The  series  is  evidently  arranged  according  to  the  ascending 
powers  of  x.  For  most  expressions,  a  slight  inspection  will  show  whe- 
ther their  development  will  contain  negative  or  fractional  exponents. 

and,  consequently,  whether  the  assumed  form  is  right.     Thus,  

1 — x~- 
cannot  be  placed  equal  to  A  +  Bx  +  Cx^,  &c.,  because  its  development 

must  contain  negative  exponents ; cannot  be  placed  equal  to 

1  +x^ 
the  assumed  form,  because  its  development  must  contain  fractional  ex- 
ponents     These  expressions,  and  others  like  them,  can,  however,  be 


UNDETERMINED    C  OE  FT  IC  I  E  N  T  S  .  341 

espauded  into  a  series  by  changing  their  form.      Place  x~^  =  z,  then, 
1  1 


1— x-^       1  — z 

•   Let  1  —  z  =  A  +  Bz  +  Cz^  +  D2'  +  Ez'  +  &c. 

We  will  find  -— —  =  1  +  z  +  z'  +  z"  +  z'  +  kc 

Now,  replace  z  by  its  value,  and  we  have ^  =  1  -f  x~^  +  x~'' 

+  x-^  +  X-'  +  &c. 

i  11 

In  like  manner,  place  x'~  =  z.     Then, = ■  =  A  +  B~ 

1  +  x-^       ^+' 
-\-  C-s^  -f  Ds^  +  E2'*  4-  &c.     We  will  find  by  the  method  of  undetermined 

1 

1  +^ 

1  J|,  3 

value,  we  have j-  =  1  —  x'~  +  x  —  x'-  +  x'^  +  &c. 

1  +  a-'^ 


FAILING    CASES. 

356.  Any  fraction,  all  of  whose  exponents  are  positive  and  entire  iu 
the  letter  according  to  which  the  series  is  to  be  arranged,  can  be  de- 
veloped, provided  the  denominator  contains  a  constant.  Again,  any 
fraction,  all  of  whose  exponents  arc  positive  and  entire  in  the  letter 
according  to  which  the  series  is  to  be  aiTanged,  can  be  developed ;  pro- 
vided, that  there  is  one  terra,  at  least,  in  the  denominator,  the  expo- 
nent of  whose  arranged  letter  is  equal  to  or  less  than  the  exponent  of 
the  same  letter  in,  at  least,  one  term  of  the  numerator.  A  fraction 
that  fulfils  either  of  the  foregoing  conditions  can  always  be  developed, 
and  the  development  will  be  the  same  as  the  quotient.  The  reason  of 
this  is  plain  ',  the  method  of  undetermined  coefiicients  gives  a  develop- 
ment for  fractions,  which  is  the  same  that  would  be  obtained  by  actual 
division,  beginning  with  the  lowest  power  of  the  arranged  letter.  Now, 
a  fraction  that  fulfils  either  of  the  foregoing  conditions,  will,  when  ex- 
panded by  division,  necessarily  give  positive  and  entire  exponents  in 
the  arranged  letter,  when  the  division  has  been  begun,  with  the  nume- 
rator and  denominator  arranged  according  to  the  lowest  power  of  this 
letter. 

Division  would  give  us  two  quotients  for zr,  according  as  we 

X  -\-  L 

made  x,  or  1,  the  first  term  of  the  divisor;  but  the  method  of  undeter- 
29* 


342  UNDETERMINED    COEFFICIENTS. 

mined  coefficients  gives  but  one  development,  whicli  is  the  same  as 
that  obtained  by  arranging  the  numerator  and  denominator  with  refe- 
rence to  the  lowest  power  of  x.  Therefore,  the  formula,  which  is  ap- 
plicable only  to  positive  and  entire  exponents,  must  fail,  when  the  least 
exponent  of  x  in  the  numerator  is  less  than  the  least  exponent  of  x  in 

.x^  4-  c^ 
the  denominator.     Thus,  '—^ cannot  be  developed,  because  o^,  which 

involves  the  lowest  power  of  x  in  the  numerator,  gives  a  quotient 
affected  with  a  negative  exponent  when  divided  by  x,  the  correspond- 
ing term  of  the  denominator. 


GENERAL   EXAMPLES. 

1.  Develop  \  -{■  x.  An$.  1  -f  x. 

2.  Develop  = ^  into  a  series. 

Am.  1  —  X  -{■  X?  —  x^  -f  cc®  —  x}  -{■  x^  —  &c. 

] ,5^ 

3.  Develop  ^j -r-  into  a  series. 

Am.  \—x  —  ^x^  —  16,x^  —  64cc''  —  256a;5  —  1024a;«  —  &c. 


1  ^2 

4.  Develop into  a  series.  .        ., 

^  1  -f-  a;  Ans.  1  —  X. 

x-'- 

5.  Develop  = ^  into  a  series. 

^1  —  X  ' 

Ans.  x-^  +  X-'  +  ^~'  +  ^"'  +  ^~'°  +  &c. 


X 

6.  Develop  '■ — -  into  a  series. 

l—xT' 

1  ~  4  5 

Am.   x'^  -\-x^  -f  a;  +  o;^  -f  x^  +  a-^  -f  &c. 


1  +  a; 
7.  Develop  :j ^  into  a  series. 


Am.  1  -f  3x  +  6a;2  +  Vlx^  -f  24.x^  +  48x5  -f  &c. 
Am.  1  —  X  +  ic"  —  x^  -f  X*  —  x^  +  x'-'  —  x'^  +  &c. 


8.  Develop  ^  _^  ^  ^  ^2  ^  ^  i'^to  a  series 


UNDETERMINED    COEFFICIENTS.  343 

9.  Develop  z, s ^ -,  into  a  series. 

Ans.  1  —  X  -i-  x'  —  x^-\-x'°  —  x''+x''  —  x'^  +  x'°  —  x"'  -^  &e. 

X^  -rS 

10.  Develop -.  Am. 


a  +  0  a  -\-  0 

x^ 

11.  Develop  — — r-  into  a  series. 

X  +  1 

Ans.  x^  —  x^  ■\-  X*  —  x'  +  x^  —  x'  +  &c. 

x^  -\-  x^  . 

12.  Develop  '- ^  into  a  series. 

x"       4.r.'        4.7:^        4x'        , 
0        2o        12o       b2o 

13.  Develop  ' —  into  a  series  arranged  according  to  the  powers 

X       y 

of  y.  Ans.  a;"-'  +  x'-'^y  +  x"'"'/  +  x'-^y^  +  .  .  .  .  y». 

14.  Develop  -^— — .  . 

x^  -\-  X  Ans.  X. 


_    ^      ,      .r^  +  3x2  +  3x  +  1  .  ^ 

15.  Develop = into  a  series. 


Ans.  x^  +  2x  +  1. 


X'' 

16.  Develop    -5 into  a  series 

x^  +  X 


Ans.  X*  —  x'  +  x''  —  x^  +  X®  —  x'  +  &c. 

xV 
—  into  a  series. 

Ans.  X  +  x'  —  x*  +  X'  —  X®  +  x^  —  x^  +  &c. 

x^  +  x^  +  X'  . 

— 5 into  a  series. 

x^  +  X 

Ans.  1  +  cc''  —  x^  +  2x''  —  2x^  +  2x«  —  2x'  +  2x^  —  2x'  +  &c. 

velop  — ; — :— - — -.  into  a  series. 

X  +  x^  +  x^  +  x" 

Ans.  l  —  x"  +  x^  —  x'  +  2x«  —  3x"  +  2x   +  x^  _  5x'". 

,       x^*  +  3x2  -\ 
velop  ~ 

x^  -+ 

The  process  fails.     Why 


X*  -{.  x^  4-  x^ 
17.  Develop  ' ^ ^ —  into  a  series 


X  4-  x^  4-  x''  +  X' 
18.  Develop j— into  a  series 


OA    -n       1       ■'?^'  +  3x2  +  3x  +  1  . 

20.  Develop into  a  series. 

x'=  +  X  Ans.    <x>. 


.>44  UNDETERMINED  COEFFICIENTS. 

1  _|_  7.x  -f  01?  . 

21.  Develop  ■ — ^ — into  a  series. 

^         1  -{-  X 

Aus.   1  +  6a:  —  5x^  +  5x^  —  5x*  +  5.r-'  _  bx^  +  &c. 

r^ a^ 

22.  Develop  '- -. 

X  —  a  A71S.   x  -\-  a. 

a'^ a^  . 

23.  Develop ^  into  a  series. 

SC-*  -f  x~^  +  2  . 

24.  Develop  ^ 1^ into  a  series. 

Ans.  2  —  x-'  +  2x~'  —  2x-^  +  2x-'  —  2a;-"'  +  &c. 
Make  x~^  =  z.     Develop,  and  replace  z  by  x~'^. 

Remarhs. 

357.  It  will  be  observed  that,  in  most  of  the  preceding  examples,  any 
term  of  the  series  after  a  certain  number  from  the  left,  can  be  formed 
from  the  term  immediately  preceding,  or  from  two  or  more  preceding 
terms,  according  to  some  fixed  law.  Thus,  in  Example  2,  every  alter- 
nate term  of  the  series  is  formed  from  that  which  precedes  it,  by  multi- 
plying by  — .T.  The  series  in  5  and  6  are  formed  in  the  same  way,  the 
multipliers,  or  all  the  terms,  being  x~'^  and  x\.  In  7,  every  term  after 
the  second  is  formed  from  that  which  precedes  it,  by  multiplying  by  2x. 
In  8,  we  may  regard  the  constant  multiplier  as  —  cc,  and  omit  every 
alternate  set  of  two  terms.  In  9,  the  multiplier  may  be  regarded  as 
—  x,  and,  when  three  terms  occur  togethei*,  these  are  omitted.  In  11 
and  16,  the  multiplier  is  —  x.  In  12,  the  multiplier,  for  all  terms  after 
the  second,  is  —  \x.  In  13,  the  multiplier  is  x~'^y.  In  17,  the  con- 
stant multiplier,  after  the  second  term,  is  —  x.  In  18,  the  multiplier, 
after  the  fourth  term,  is  —  x.     In  19,  there  is  no  law  of  formation. 

When  the  terms  of  a  development  are  formed  from  those  which  pre- 
cede according  to  some  fixed  law,  the  development  is  called  a  recurring 
series.  Multiplication  is  not  the  only  mode  of  forming  the  succeeding 
terms;  sometimes  they  are  formed  by  addition  or  subtraction,  and  even 

by  a  combination  of  two  of  these  methods.     Thus,    ; ^ —^ 

expanded,  becomes  1  +  a;  +  2^^  4-  4x^  +  Ix^  -\-  13a;^  -|-  24x«  +  &c. 


UNDETERMINED    COEFFICIENTS.  o45 

The  literal  parts  are  formed  by  multiplying  by  x,  the  first  two  terms  have 

the  same  coefficient,  and  each  succeeding  coefficient  is  equal  to  the  sum  of 

1  4-  2a; 
the  three  which  precede  it,  &c.     Again,  ^ ^^— — ^  =  1  +  3a;  +  4x* 

+  Ix"  +  lla=^  +  18a;5  +  29:r«  +  47x^  +  &c.,  in  which  the  coeffi- 
cients after  the  second  are  equal  to  the  sum  of  the  two  preceding 
coefficients. 


PARTICULAR   CASES. 

358.  A  fraction  involving  irrational  monomials  may  be  treated  as  in 
the  following  example. 

Required,  the  development  of  -^ ^.    Let  x  =  z^.    Then,  ai  = 

2^,  x'l  =  z"^,  z  ^  xl.     Then,  the  fraction  becomes  ~, 3,  and,  by  the 

method  of  undetermined  coefficients,  .j .  =  A  +  Bz  +  Csr^  4-  Ds' 

1  —  2r 

+  Ez^  +  Fs^  +  &c.  From  which  we  get  A  =  0,  B  =r  0,  C  =  0, 
D  =  1,  E  =  —  1,  F  ==  0,  G  =  1,  H  =  —  1,  &c.  Hence,  ^^=4 
z=^  —  z'^  -\-  z^  —  z'  +-^  —  2'°  4-  &c,  and,  by  replacing  z  by  its  value, 

we  have    \  ^xl  —  |  -\- x  —  x].  +  a-|  —  a-J  +  &c. 

i  —  x^ 

Since  all  examples  of  the  same  kind  may  be  treated  in  the  same  way, 

we  derive  the 

RULE. 

Place  the  variable  equal  to  a  new  variable,  affected  with  an  expo- 
nent equal  to  the  least  common  multiple  of  the  denominators  of  the 
exponents  of  the  old  variables.  Develop  the  neto  fraction,  and  replace 
the  neio  variable  by  its  vahie  in  terms  of  the  old. 


1,  Expand    -y ^  into  a  series. 

Ans.  x\  —  xf  4-  ^g  —  ^1  4-  &c. 

2.  Expand    '  *       ^^.  Ans.  x\  —  a;l  +  x  |  —  x%  +  &c. 

1  4-  a;  b  b  4 


346  UNDETERMINED    C  0  E  F  FI  CI  E  N  T  S  . 

859.  Fractions  involving  parentheses,  differing  only  in  their  expo- 
nents, may  be  treated  as  the  following. 

Required,  the  development  of 


Let  1  +  X  =  z^.     Then,  ^^    ,    ^, ^-,    ,     ^  = -,  which,  when 


(l+aOi  — (1  +:r)- 
X  _  ^'  —  1 

developed,  gives  z  +  z^  +  z'^  +  z*  -\-  &c.     Now,  replace  z  by  its  value, 
andwehave^^_^   ^^,^^^_^^^    =    (1    +  ^)J    +    (1    +   .)>    -f 

(1  +  cc)i  +  (1  +  ^y  +  &c. 


RULE. 

Place  the  common  parenthesis  equal  to  a  Jietv  variable,  raised  to  a 
power  deviated  hy  the  least  common  multiple  of  the  denominators  of  the 
p>arentlieses,  and  proceed  as  hefore. 


EXAMPLES. 

1.  Expand    -^ ^- — ^-—^ ^J^to  a  series. 

{l  —  x)}^  —  {l  —  xy\ 

Ans.  (1  —  x)i  +  (1  —  x)\  +  (1  —  xf  +  (1  —  X)  ]  +&C. 

2-  Expand    ^_^^_^,  into  a  series. 
Ans.  (1  — a;)f  — (1  — x)+(l  — x)--4  — (1— a-)-|+(l— a;)-2  — &c. 

m 

Expressions  of  the  form,  ( — j— 7-)  ,  iiiay  be  expanded  by  placing 
a  —  bx  =  2°,  deducing  the  value  of  a  -f  bx,  and  proceeding  as  before. 

Take,  (l^^\^.      Place,  1  —  x  =  z'.      Then,  x  =  1  —  z^,  and 

'\1  +  x/  ' 

,         XT  /l—X\i  Z^  Z^ 

l  +  x=2-z\     Hence,  (^-^)     =  43347^^76  =  ^=4^+4- 
Develop  this  fraction  and  replace  z  by  its  value  (1  —  cc) J. 

360.  It  is  obvious  that  there  are  an  infinite  number  of  irrational 
expressions  that  may  be  developed  by  first  making  them  rational  in 
terms  of  a  new  variable,  expanding  the  new  expression,  and  replacing 
the  new  variable  by  its  value  iu  terms  of  the  old.  So,  there  are  an 
infinite  number  of  expressions,  involving  negative  exponents,  that  may 


DERIVATION.  347 

be  expanded  by  substituting,  for  the  old  variable,  a  new  variable  affected 
with  a  positive  exponent,  and  proceeding  as  before.     Thus,  let  it  be 

required  to  expand  -^j ^.    Let  x~'^  =  z.    Then,  x~*  =  z^.    Hence, 

jj-^  +  x-'  —  x-'  +  x-"'  —  x-''+&c. 

There  are  many  expressions,  however,  involving  negative  and  frac- 
tional exponents,  that  cannot  be  developed  by  the  method  of  undeter- 
mined coefficients. 


DERIVATION. 

361.  When  one  quantity  depends  upon  another  for  its  value,  it  is 
said  to  be  a  function  of  the  quantity  upon  which  it  depends.  The 
dependent  quantity  is  the  function,  and  that  upon  which  it  depends  is 
called  the  variable.  Thus,  in  the  equation,  y  =  2x,  y  is  the  function 
and  X  the  variable,  because,  by  changing  the  value  of  x,  y  may  be  made 
to  have  any  system  of  values.  Thus,  when  a;  =  0,  *,  1,  2,  3,  &c  j 
y=0, 1,2,4,  0,  &c.,  it  is  evident  that,  in  every  single  equation  involving 
two  unknown  quantities,  there  is  a  function  and  variable.  Either  of  the 
unknown  quantities  may  be  assumed  at  pleasure  as  the  function,  because 
both  unknown  quantities  are  mutually  dependent  upon  each  other.  But 
it  is  usual  to  regard  that  unknown  quantity  as  the  function,  with  refer- 
ence to  which  the  equation  has  been  solved.     If  the  equation,  y  =  2a", 

is  solved  with  respect  to  a;,  we  have  a;  =  -^.     Then,  x  is  the  function 

a 

and  y  the  variable,  and,  by  assuming  arbitrary  values  for  y,  x  may  be 
made  to  have  an  infinite  system  of  values.  The  most  general  form  of  a 
solved  equation  of  the  first  degree  with  two  unknown  quantities  is,  y  = 
ax  -f  b.  In  this,  a  and  b  are  called  constants,  because  their  values 
undergo  no  change.  It  is  plain  that  y  depends  for  its  value  upon  x,  a, 
and  b,  but,  as  its  value  only  changes  with  x  (since  x  is  the  only  variable), 
7j  is  called  a  function  of  x  only.  If  we  have  a  single  equation  involving 
three  unknown  quantities,  any  one  of  these  unknown  quantities  is  a 
function  of  the  other  two. 

If  we  have  y  =  ox,  ov  y  =  ax  +  b,  and  x  —  viz,  or  x  =  viz  -f  v. 


348  DERIVATION. 

we  call  y  an  explicit  fuuetion  of  x,  and  an  implicit  function  of  z, 
because  y  directly  depends  upon  x  for  its  value,  and  indirectly  upon  z. 
It  is  plain  that  we  may  have  explicit  functions  of  any  number  of 
variables,  and,  also,  implicit  functions  of  any  number  of  variables.  We 
will,  however,  confine  ourselves  to  explicit  functions  of  single  variables. 

If  we  take  the  equation,  y  =  ax  +  h,  and  attribute  to  x  arbitrary 
values,  0,  1,  2,  3,  4,  &c.,  and  call  the  corresponding  values  of  y,  y,  y', 
y",  y"',  &c.,  we  will  have  y  =  h,  y'  =  a  +  h,  y"  =2a  -\-  b,  y"  =  3a 
+  h.  Then,  y'  —  y  =  a,  y"  —  ?/'  =  «,  y'"  —  y"  =  a,  &c.  That  is, 
the  difierence  between  any  two  consecutive  states  of  the  function  is 
constant.  This  constant  increase  of  the  state  of  the  function  is  called 
the  differential  or  derivative  of  the  function.  It  is  evident  that,  when 
the  function  and  variable  are  both  linear,  that  is,  both  of  the  first 
degree,  the  difference  between  two  consecutive  states  of  the  function 
will  always  be  constant,  and  then  will  be  truly  the  derivative  of  the 
function.  Thus,  take  the  equation,  y  =:  2x  -\-  2,  and  let  x  have  the 
constant  increment  J,  beginning  at  0 ;  that  is,  let  a;  =  0,  0  +  i,  J  +  ^, 
1  +  i^j  I2  +  J;  then  the  difference  between  two  consecutive  values  of 
y  will  always  be  constant.  For,  we  have  y  =  2,  y'  =zl  +  2  =  3, 
y"  =  2  +  2  =  4,  y'"  =  3  +  2  =  5,  &e.,  and,  consequently,  y'  — y  = 
l=y"—y'=  y'"  —  y",  &c. 

But,  when  the  variable  is  of  the  second  degree,  and  the  function  of 
the  first,  the  difference  between  two  consecutive  states  of  the  function, 
corresponding  to  constant  increments  of  the  variable,  will  not  be  con- 
stant. Thus,  take  the  equation,  y  =  x^,  and  let  x  have  the  constant 
increment,  1,  beginning  at  zero.  Then,  y  =  0,  y'  =zl,  y"  =  4,  y'"  =  9, 
&c.,  and  y' — ^=1?  y" — y' =^^j  1/'" — y  =  5.  The  student  of 
geometry  will  understand  that  there  ought  to  be  no  constant  increment 
to  the  function,  corresponding  to  a  constant  increment  to  the  variable  in 
the  equation,  y  =^  0?.  For,  y  expresses  the  surface  of  a  square  of  which 
X  is  the  side,  and  the  surface  of  course  increases  more  rapidly  than  the 
side.  If  the  function  is  of  the  first  degree,  and  the  variable  of  the 
third,  it  will  be  seen  that  the  difference  between  two  consecutive  states 
of  the  fuuetion  varies  still  more  widely  from  a  constant.  The  function 
and  variable  must  then  both  be  of  the  first  degree,  in  order  that  the 
difference  between  two  consecutive  states  of  the  function  may  be  con- 
stant. But,  to  return  to  the  illustration  of  the  square,  it  is  plain  that, 
if  the  constant  increment  to  the  side  of  the  square  was  indefinitely 
small,  the  difference  between  two  consecutive  states  of  the  function, 
which  represents  the  increment  of  the  fuuetion,  would  be  so  nearly  con- 


DERIVATION.  349 

stant  that  it  might  be  regarded  as  constant.  Suppose  y  =  a:^  and  that 
x=^\  foot.  Then  the  surface  of  the  square  is  one  square  foot.  Now, 
suppose  the  side   of   the  square  to  receive   the    constant   increment, 

1  2  1 

j-^    part   of  a   foot.     Then,   y  ==  1,  y  =  1  +  ^-^-^  +  ^j^^^, 

4  4  2  1 

y"  =  ^^  10000  +  (TOOOO/'  *^''-    ^^^°'  •^' "^  "^  10000  +  (10000?' 

2  2 

?/"  —  7/  = 1 -„.    And  we  see  that  the  diflPerence  between 

■^        ^        10000  ^  (10000)'' 

the  second  and  first  states  differs  only  from  the  difference  between  the 
third  and  second,  by  jq^oo- 

Upon  this  principle  the  derivative  or  increment  of  a  function  is 
found.  It  is  the  difference  between  two  consecutive  states  of  the 
function,  when  these  states  are  indefinitely  near  to  each  other,  and  it 
expresses  the  increment  of  the  function.  The  word  increment  is  used 
in  its  algebraic  sense.  When  any  state  of  the  function  is  less  than  the 
preceding,  the  derivative  is  truly  a  decrement. 

To  find  the  derivative  of  the  function,  y  =  a-^,  let  h  represent  the 
infinitely  small  constant  increment  to  x.  Then,  y  =  x^,  y'=  (x  +  h)^ 
=  x"^  +  2x7*  +  h^,  and  y'  — y  =^  2xh  -f  h^.  Now,  since  h  is  infinitely 
small  by  hypothesis,  h^  will  be  an  infinitely  small  quantity  of  the  second 

order,  and  may  therefore  be  neglected.     Thus,  -z — rrp—   is  very  small, 
'  •'  °  '1  million  •' 

but  -rz, 777^ — r%  is  much  smaller.    Now,  by  making  h  indefinitely  small, 

(1  million/  7    .;  o  J  ? 

we  have  taken  the  states  indefinitely  near  to  each  other,  and,  conse- 
quently, y'  —  y  is  the  derivative  of  y  =  x?.  Hence,  the  derivative  of 
0^  is  2x7t.  It  is  usual  to  represent  h  the  increment  of  the  function  by 
dx,  read  differential  of  x.  The  derivative  of  the  square  of  any  variable 
function  is  then  twice  the  first  power  of  the  variable  into  the  differential 
of  the  variable.  The  differential  coefficient  of  a  function  is  the 
differential  of  the  function  divided  by  the  increment  of  the  variable, 

and  is  generally  expressed  by  '^— — '- .     In  the  preceding  example  the 

differential  coefficient  of  x^  is  2.r ;  and,  in  general,  knowing  the 
differential  of  the  function,  we  get  the  dift'erential  coefficient  by  dividing 
by  the  increment  or  differential  of  the  variable ;  and,  conversely,  we  get 
the  differential  of  the  function  from  the  differential  coefficient,  by  mul- 
tiplying by  the  differential  of  the  variable.  It  will  be  seen  that,  when 
30 


350  DERIVATION. 

wc  make  h  =  0,  or  indefinitely  small,  that  the  differential  coefficient 
assumes  the  form  of  ^,  for  then  i/'  =  y. 

To  find  the  differential  coefficient,  we  have  the  following 


RULE. 

Give  to  the  variahle  a  varialle  increment,  h,  and  find  the  neiv  state 
of  the  function.  Take  the  difference  between  the  new  and  old  states  of 
the  function,  and  reject  the  terms  involving  the  higher  powers  of  the  in- 
crement, as  heing  infinitely  small  quantities  of  the  second,  third,  &c., 
orders.  Next  change  h  into  dx,  dy,  or  dz,  &c.,  according  as  the  va- 
riahle is  X,  J,  or  z,  &c.  We  then  have  the  differential  of  the  function. 
Divide  the  differential  of  the  function  hy  the  differential  of  the  va- 
riahle, and  we  have  the  differential  coefficient. 

Required  the  differential  coefficient  of  the  function,  y  =  x^.  By  the 
rule,  we  have  y'  =  (x  +  hy  =  x^  +  3x-/i  +  oh^x  +  7/,l  Hence, 
//'  —  y  =  Sx^h  +  Sh^x  -}-  A"  =  SxVi,  or  Sx^dx,  when  h  is  infinitely 

small.     Therefore,  ^~"~ '^  =  3.x-l 
h 

Newton  regarded  all  algebraic  expressions  as  the  representatives  of 
lines,  surfaces,  or  solids;  and  supposed  lines,  whether  straight  or  curved, 
to  be  generated  by  the  flowing  of  points  according  to  fixed  laws,  surfaces 
to  be  generated  by  the  flowing  of  lines,  and  solids  to  be  generated  by 
the  flowing  of  surfaces.  Thus,  a  point,  moving  or  flowing  according  to 
the  law  that  it  shall  always  be  in  the  same  plane,  and  at  the  same  dis- 
tance from  a  point,  will  generate  the  circumference  of  a  circle;  and  a 
straight  line,  flowing  with  the  two  conditions  of  being  constantly  parallel 
to,  and  at  the  same  distance  from  a  fixed  line,  will  generate  the  surface 
of  a  cylinder.  A  straight  line  flowing  in  the  same  direction  with  the 
condition  that,  in  all  its  positions  it  shall  continue  parallel  to  itself  in 
its  first  position,  will  generate  a  plane.  One  line  revolving  around 
another,  to  which  it  is  perpendicular,  _L,  will  generate  the  surface  of  a 
circle.  A  flowing  square  will  generate  a  cube;  a  flowing  semicircle,  a 
sphere.  Newton  regarded  the  increment  of  the  function  as  expressing 
the  uniform  rate  of  increase  of  the  function,  and  called  it  fiuxion. 
The  thing  generated  he  c^W^di  fluent.  It  is  evident  that  the  fluxion 
does  not  express  the  uniform  rate  of  increase,  except  when  the  states 
are  taken  indefinitely  near  to  each  other.  Suppose  a  cannon-ball  to 
leave  the  mouth  of  a  piece  with  a  velocity  of  2000  feet  per  second,  and 
that  this  velocity  is  reduced  to  1200  feet  per  second  at  the  end  of  the 


DERIVATION.  351 

third  second.  It  is  plain,  that  the  diiFerence  between  the  spaces  passed 
over  in  any  two  consecutive  instants  of  time  will  not  be  equal  to  the 
difference  between  the  distances  passed  over  in  any  other  two  consecu- 
tive instants,  unless  those  instants  are  inappreciably  small.  But,  for 
the  millionth  part  of  a  second,  the  velocity  might  be  regarded  as  con- 
stant. If  the  instant  was  then  taken  thus  small,  the  difference  between 
the  spaces  in  two  consecutive  instants  would  be  constant. 

The  differential  of  the  function  has,  in  accordance  with  the  New- 
tonian theory,  been  defined  to  be  the  uniform  rate  of  increase  or 
decrease  of  the  function.  The  differential  of  a  constant  then,  must  be 
zero,  since  a  constant  admits  of  neither  increase  or  decrease. 


Theorem  I. 

362.  The  differential  of  the  algebraic  sum  of  any  number  of  func- 
tions of  the  same  variable  is  equal  to  the  algebraic  sum  of  their  diffe- 
rentials. 

Let  ?/=-±iudczvzt:w±z;  «,  v,  lo,  and  z,  being  functions  of  the 
same  variable,  x.  Give  to  the  variable  a  variable  increment,  h,  and 
call  the  new  states  of  the  function  «',  v',  xo',  and  z.  Then  y  =  «'  =h 
v'  zh  w'  ±  z',  and  y'  —  1/  =  u'  —  m  =fc  (v'  —  v)  ±  (ic'  —  ic)  ±  (/-^  z) 
as  enunciated,  since  u'  —  «  is  the  differential  of  m,  v'  —  v,  the  diffe- 
rential of  V.  A  shorter  notation  is  dy  =  du  rt  dv  rfc:  dw  ±  dz,  and  is 
read  differential  of  y,  equal  to  the  differential  of  ?<,  plus  or  minus  the 
differential  of  r,  &c. 

Let  y  =  ax^  +  ex.     Then,  dy  =  2axdx  -f-  cdx. 

Theorem  II. 

363.  The  differential  of  the  product  of  a  constant  by  a  variable,  is 
equal  to  the  product  of  the  constant  by  the  differential  of  the  variable. 

Let  y  =  Ax.  Then,  dy  =  Adx.  For,  y'  =  A  (x  -{-  h),  and 
y' — y  =  Ah,  or  dy  =  Adx. 

Let  u  =  my.     Then,  du  =  mdy. 

Theorem  III. 

364.  The  differential  coeflacient  of  the  power  of  a  quantity  is  equal 
to  the  exponent  of  the  power  into  the  quantity  affected  with  an  expo- 
nent less  by  unity  than  the  primitive  exponent. 


352  DERIVATION. 

Lety  =  a;".     Then,  ^^^- =  «x"-'. 

For,  give  to  x  a  variable  increment,   h ;  then,  y'  =  (x  ■{■  hy,  and 

y  —  ?/ (a?  +  A)  "  —  a;"       (x  +  7i)  °  —  x"       «"  —  x" 

h       ~  h  ~    (x  +  A)  —  X    ~    a  —  a;  '    "^      -P    ' 

iuij  X  +  7t  by  a.     But,  ■ =  «°~'  +  «°~^j:;  +  (/""'x^  +  a^'^x^  + 

a  —  X 

&c.,  on  to  n  terms.     Now,  when  h  is  made  indefinitely  small,  x  is  equal 

to  a,  the  first  member  of  the  equation  in  y  then  becomes  equal  to  the 

diflferential  coefficient,  and  the  second  member  becomes  a"~'  -f  a"~'  + 

«"~'  +  &c.  =  ««""',  as  enunciated. 

Hence,  we  have  ---  =■  ??«""',  and,  consequently,  dy  =  na^~hlx.    The 

differential,  as  well  as  the  differential  coefficient  of  a  powei-,  is  then 
known. 

Theorem  IV. 

365.  The  differential  of  the  product  of  two  functions  of  the  same 
variable  is  equal  to  the  sum  of  the  products  which  arise  from  multiply- 
ing each  function  by  the  differential  of  the  other  function. 

Let  y  =1  uv.  Then  dy  =  udv  •-(-  vdu,  in  which  u  and  v  arc  both 
functions  of  the  same  variable,  x;  the  new  state  of  the  function  is 
n'  =  u  +  du,  and  the  new  state  of  the  function  v'  =  v  +  dv.  Hence, 
y'  =  u'v'  =  (u  -\-  du)  (v  +  dv)  ^uv  -{•  udv  +  vdu  +  diidv.  But, 
since  du  and  dv  are  indefinitely  small,  their  product,  dudv,  is  an  in- 
definitely small  quantity  of  the  second  order,  and  may  therefore  be 
neglected.  Hence,  y'  =  uv  +  udv  +  vdu,  and  y' — y,  or  dy  =  udv 
+  vdu,  as  enunciated. 

Corollary. 

366.  1st.  The  differential  of  the  product  of  any  number  of  functions  of 
the  same  variable  is  equal  to  the  sum  of  the  products  which  arise  from 
multiplying  the  differential  of  each  function  by  the  product  of  the  other 
functions. 

Let  y  =  swuz,  in  which  s,  w,  %i,  and  z  are  functions  of  the  same 
variable,  z.     Then,  dy  =  icuzds  +  suzdw  -f  swzdu  -f  sicudz. 

To  show  this,  we  will  first  get  the  differential  of  the  product  of  three 
variables.  Let  y  =  szv.  Make  sz  =  u.  Then  y  =  uv,  and,  from  what 
has  just  been  shown,  dy  =  udv  +  vdu  =  szdv  -f  vd  (sz)  =  szdv  -f- 
V  (sdz  -\-  zds)  =  szdv  -f  vsdz  -\-  vzds,  as  enunciated. 


DERIVATION.  853 

Now,  knowing  the  differential  of  the  product  of  three  variable  func- 
tions, we  can  readily  pass  to  that  of  four.  For,  let  y  =  swuz.  Then, 
Ji/  z=  svmJz  +  zd  (^Rwu)  =  sicuch  -f  z  (sic(7v  -f  sudw  +  icuds)  =  swudz 
+  zswdu  -\-  zsudw  +  zicuds.  And,  it  is  plain  that  the  same  process  can 
be  extended  to  the  product  of  any  number  of  functions. 

2d.  Since  the  differential  of  the  product  of  Av  is  Adv  -f  vdA  = 
Adv,  A  being  a  constant,  we  have  Theorem  II.  demonstrated  in 
another  way. 

Theorem  V. 

367.  The  differential  of  a  fraction  is  the  denominator  into  the  diffe- 
rential of  the  numerator,  minus  the  numerator  into  the  differential  of 
the  denominator,  divided  by  the  square  of  the  denominator. 

Let  y  =  — ,  s  and    v  bciuu;   both   functions    of  .'-.     Then,    di/  = 

V 

vds  —  sdv 


For,  since  y  =  — ,  we  have  no  =  s  ;  and  yv  being  equal  to  s,  the  iu- 

V 

crement  of  yv  must  be  equal  to  the  increment  of  s.    That  is,  d  (yv)  = 

ds        ydr         ds       sdv       vds  —  sdv 
ds,  or  ydv  +  vdy  =  as,  or  i/y  =  ^  — ' = ~  = , 

as  enunciated.   In  the  expression,  "^ — ,  we  substituted  for  y  its  value,  — , 

V  V 

and  then  reduced  the  two  fractions  to  a  common  denominator. 


CoroUary. 

368.   1st.  When  the  denominator  is  constant,  dv  is  zero,  and  dy  =3 

vds  —  sdo       vds       ds  ,  .        ,.„  •  1      p     1 

=  —-T-  =  — ,   equal    to   the    dmerential    ot   the   numerator 

v^  v^  V 

divided  by  the  constant  denominator.     "We  might  arrive  at  this  result 

s  Is 

in  another  way,  for,  when  v  =  a   constant,  —  may  be  written  — ,  and, 

1  Is 

since  —  is  constant,  the  differential  of  —  will,  by  Theorem  II.,  be 

V  '  V 

Ids  ds 
'  di 
30 


,  or 
dv 


354  DERIVATION. 

369.    2(1.   When    the    numerator    is    constant,    ds  =  0,    and    dy  = 

vds  —  sdr   -  ,  ado  .  , 
,  becomes  c/y  =  —  —2}^  negative  result. 

This  ought  to  be  so,  for,  when  the  numerator  is  constant  and  the 
denominator  variable,  any  increment  to  the  variable,  x,  upon  which  v 
depends,  will  decrease  1/ ;  and,  consequently,  cTy,  which  expresses  the 
algebraic  increment  of  7/,  is  truly  a  decrement,  and  ought,  therefore,  to 
have  the  negative  sign.  In  the  case  supposed,  y  is  a  decreasing  function 
of  the  variable,  x,  and  we  see  that  the  differential  of  a  decreasing 
function  is  negative. 

We  are  now  prepared  to  find  the  differential  of  any  function  affected 
with  any  exponent,  whether  positive  or  negative,  fractional  or  entire. 


Theorem  VI. 

370.  The  differential  of  a  quantity  affected  with  any  exponent,  is 
the  product  arising  from  multiplying  the  exponent  of  the  power  into 
the  quantity  affected  with  an  exponent  algebraically  less  by  unity  than 
the  primitive  exponent,  into  the  differential  of  the  variable. 

We  will  first  suppose  the  quantity  to  be  affected  with  a  negative 

exponent.      Let  y  =  r~°  =  —^.     Then,  by  Corollary  2d,  of  the  last 

Theorem,  dy  = ^^^  = -^   =  —  nv~''~\  as  enunciated. 

Next,  suppose  the  exponent  to  be  a  positive  fraction,  ■>/  z=v\  Then, 

7/  r=  \/v'',  or  1/'  =  v'.     And,  taking  the  increments  of  both  members, 

we   have,  by  Theorem  III.,  since  r  and  s   are    positive   and    entire, 

,  ,                 -.          rv^~^dv         r        if~^dv        r         v^~'dv 
.«?/-'  dy  =  rv'-h/i',     or     di/  =  j—  =  —  X  — —7-  =  —  X  — 

r  r^~^dv  r  v~^dv  r    —  r    -i_i 

V  '  v~ 

enunciated. 

Finally,  suppose  y=  v~  ^  ■   we  will  find  by  proceeding  exactly  in  the 

:?ame  way  as  when  the  exponent  is  a  positive  fraction,  dy  ^= -  x 

s 

V   »    dv,  as  enunciated. 


DERIVATION.  355' 


Corollary. 

370.  1st.  The  differential  of  any  parenthetical  expression  is  equal  to 
the  exponent  of  the  parenthesis  into  the  parenthesis,  raised  to  a  power 
algebraically  less  by  unity  than  at  first,  into  the  differential  of  the 
quantity  within  the  parenthesis. 

This  is  evident,  since  we  may  represent  the  quantity  within  the 
parenthesis  by  a  single  variable,  v.  The  parenthetical  expression  will 
then  assume  the  form  of  v",  and  may  be  differential  according  to  the 
rale. 

Let  y  =  (a  +  hx^y,  place  a  -\-  hx^  =  /•,  then  y  =  z",  and  dy  = 
nv^-klv  =  n  (a  +  hx^y'-\l  (a  +  hx^)  =  n  (a  +  bx^)"^^  2l)xdx  = 
2bn  (a  +  hx^y-^xdx. 

372.  2d  The  differential  of  a  radical  expression  is  equal  to  the  diffe- 
rential of  the  quantity  under  the  radical,  divided  by  the  index  of  the 
radical  into  the  radical  raised  to  a  power  algebraically  less  by  unity  than 
the  primitive  index.  This  is  merely  a  particular  case  of  the  preceding, 
for  a  radical  is  nothing  more  than  a  parenthetical  expression  affected 
with  a  fractional  exponent,  the  numerator  of  the  fraction  being  unity. 

Let  -yU  be  the  radical;  this  is  equal  to  v~,  and  its  differential  then 

must  be  —  r^"'  dv  =  —  v~^  dv  =  — j^  =  — ■,  as   enunciated. 

n  n  nv—        „ ;/ i^n-i' 

"When  n  =  2,  the  radical  is  of  the  second  degree,  and  the  expression 
becomes  — =,  that  is,  the  differential  of  a  radical  of  the  second  decree 

is  equal  to  the  differential  of  the  quantity  under  the  radical  sign  divided 
by  twice  the  radical. 

EXAMPLES. 


1.  Required  the  derivative  oi  y  =  l/a  -\-  x^. 

.         ^  2xdx 

Ans.  dy  ■ 


Si/(a  +  xy 


2.  Required  the  derivative  of  y  =  v'"  +  bx'- 

.         -,  2bxdx 

Ans.   dy 


2^  a  +  bx^ 


3.  Required  the  derivative  of  "^ ax  +  ix°. 


(a  +  9iZ»x°~')  dx 
Ans.  - 


m'^(ax  -f  ftx")™-' 


356  BINOMIAL    FORMULA. 


GENERAL   EXAMPLES. 


1.  Required  the  derivative  of  «  +  ix  —  cx^  +  mx^. 

Alls,   (h  —  2cx  +  omx^)  (h\ 

2.  Required  the  derivative  of — . 


^         cx^hilx  —  (a  -f  Jjx)  2cxdx 
Ans.  ■ — 


3.  Required  the  derivative  of  cc"  (a  +  Ixy. 

Ans.  m(a  +  bxyx.'^~^  dx  +  nx"'  (a  -f  ia;)""'  hdx. 


BINOMIAL    FORMULA. 

373.  By  actual  muhiplicatiou  we  can  readily  find 

(a  +  xf  =  a''  4-  2ax   +  o:% 

(a  +  xf  =  a^  +  Sa\v  +  Sax''  +  x% 

(a  +  xy  =  a*  +  4a''a;  +  QaV  +  Ao'x^  +  x\ 

We  observe  a  simple  law  in"  regard  to  the  exponents  in  these  three 
developments.  The  exponent  of  a  in  the  first  term  is  the  power  of 
the  binomial,  and  this  exponent  goes  on  decreasing  by  unity  unto  the 
last  term,  in  which  it  is  zero.  For,  x^,  x^,  and  x*  may  be  written,  a^x^, 
a".*',  and  a°x*.  The  exponent  of  x  is  zero  in  the  first  term,  and  goes 
on  increasing  to  the  last  term,  in  which  it  is  equal  to  the  power  of  the 
binomial.  With  regard  to  the  coefficients,  we  observe  that  the  coeffi- 
cients of  the  extreme  terms  are  unity,  that  the  coefficient  of  the  second 
term  is  the  degree  of  the  binomial,  and  that  the  coefficient  of  the  third 
term  can  be  found  by  multiplying  the  coefficient  of  the  second  term  by 
the  exponent  of  a  in  that  term,  and  dividing  by  the  number  of  terras 
which  precede  the  required  term.  Thus,  in  the  development  of 
(a  +  xy,  to  find  this  coefficient,  multiply  4,  the  coefficient  of  the  second 
term,  by  3,  the  exponent  of  o  in  that  term,  and  divide  by  2,  the  number 
of  terms  which  precede  the  required  term,  the  quotient  6  is  the  co- 
efficient of  the  third  term.  So,  to  find  the  coefficient  of  the  fourth  term 
of  (a  +  xy,  multiply  G,  the  coefficient  of  the  third  term,  by  2,  the  ex- 
ponent of  a  in  that  term,  and  divide  the  product,  12,  by  3,  the  number 


BINOMIAL    FORMULA.  357 

of  terms  whicli  precede  the  required  term,  the  quotient,  4,  is  the  co- 
efficient of  the  fourth  term.  This  remarkable  law  in  regard  to  the  co- 
efficients, holds  true  even  for  the  second  and  last  terms. 

We  observe,  furthermore,  that  the  sum  of  the  exponents  of  a  and  x 
in  every  term  is  equal  to  the  exponents  of  the  binomial  and  that  the 
coefficients,  at  equal  distances  from  the  extremes,  are  equal. 

Newton  showed  that  the  law  in  regard  to  the  exponents  and 
coefficients  was  general  for  any  binomial  of  the  form  (a  +  x)"",  and 

that  we  would  have  (a  +  x)'"  =  a"  +  ma'"-^x  -i ^^ ^-^ \- 

m  (m  —  1)  (m  —  2)  a'^^x^ 

2T3  + ^  ■ 


DEMONSTRATION. 

374.  To  demonstrate  this  useful  and   remarkable    Theorem,  let  us 

assume  (a  -f  x)"'=A  -f  A'x  -f  A'V  +  A''^  + A'"x"' ;  in  which 

m  is  taken  positive  and  entire,  and  A',  A",  &c.,  independent  of  x. 
Then,  by  the  principle  of  undetermined  coefficients,  we  have  a  right  to 
make  a;  =  0  in  both  members  of  the  assumed  identical  equation. 
Making  a:  ^  0,  we  have  A  =  a". 

It  was  shown  that,  when  two  functions  of  the  same  variable  were  ecjual, 
that  their  diffiDreutials  were  equal.  Hence,  taking  the  derivations  of 
both  members  of  the  assumed  equation,  we  have  ?»(a  +  j)'"~hlx  = 
(zV  +  2A"a;  -f  3A'V  -f  &c.)'/^,  or,  dividing  by  (Jx,  m(a  +  j-)"-'  = 
A'  -f  2A"a;  -f  ^A"'x^  +  &c.  (N).  Now,  make  x  =  0,  and  we  get, 
ma"""'  =  A'.  Again,  taking  the  derivatives  of  both  members  of  (N), 
and   dividing    by   dx,    we    get,    m  (m  —  1)   (a  +  a-)""^  =  2A"  + 

2.3.  A"'x  +  &c.  (P).     Making  x  =  0,we  have  A"=  "^  ("^  —  1)  """'^ 

Again,  taking  the  derivation  of  (P),  dividing  by  dx,  and  making  a:  ^  0, 

there  results  A     =  — ^ o     o. •     -In  hke  manner,  we  can 

find  A""  =  ^»0»'— 1)C»'  —  -)  ('«  —  3) ^"-^ 


2.3.4 
And  A"  will  plainly  be  equal  to 

m  (m  —  1)  (m  —  2)  ....  (m  —  (m  —  1)  )  a"-"  _ 
1.2.3....  {m  —  2)  (m  —  1)  m  ' 

and,  since  the  factors  of  the  denominator  are  the  same  as  those  of  the 
numerator,  only  written  in  reverse  order,  we  have  A"  =  1. 


358  BINOMIAL     FORxMULA. 

Now,   replace    A,   A',   A",  &c.,    by  tlielr   values    iu    the    assumed 
identical    equation,    and    we    have   (a  +   x)-  =  d'"  +  ma'"-^  x  + 

m  {m  —  1)  ff™""^  X-  m  (m  —  1)  (m  —  2)  n"'~^  x^ 


1.2  '  1.2.8 

m  (to  —  1 )  («i  —  2)  .  .  .  {m  —  n  +  2)  0"--"+'  x<^-' 

■  1  .  2  .  3  .  4  .  .  .  (n  —  1)  "^ ^  • 

The  term, 

m  (m  —  1)  (:m  —  2)  .  .  .  (»t  —  n  +  2)  an>-n+'  x"-' 

1  .  2  .  3  .  .  ,  (ri  _  1)  ' 

is  called  the  n*"",  or  general  term.  An  inspection  of  the  formula  will 
show  that  the  first  factor  of  the  numerator  of  the  coefficient  of  a  iu 
any  term  is  m,  the  next  factor  (m  —  1),  and  so,  on  decreasing  by  unity, 
unto  the  last  factor,  which  is  m,  diminished  by  two  less  than  the  place 
of  the  term.  Thus,  the  last  factor  of  the  numerator  of  the  coefiicient 
of  the  4tli  term  is  (m  —  2).  So,  the  last  factor  of  the  numerator  of 
the  coefficient  of  the  n^^  term  must  be  (m  —  n  +  2).  The  denominator 
of  the  coefficient  of  a,  in  any  term,  is  always  the  continued  product  of 
the  natural  numbers,  from  1  up  to  one  less  than  the  place  of  the  term. 
Hence,  the  denominator  of  this  coefficient  in  the  n^^  term  must  be 
1 .  2  .  3  . .  .  .  (n  —  1).  In  regard  to  the  exponents,  we  see  that  a,  in  any 
term,  is  always  affected  with  the  exponent  of  the  binomial,  diminished 
by  the  number  of  the  term  less  one,  and  the  exponent  of  x,  in  any 
term,  is  always  one  less  than  the  place  of  the  term.  Hence,  for  the  ?i*'' 
term,  we  have  «"'-''+'  a;"~'. 

The  binomial  formula  is  usually  written  iu  the  ascending  powers  of  «, 
instead  of  x.  To  develop  (x  +  «)"  in  the  powers  of  a,  it  will  be 
necessary  to  regard  a  as  the  variable,  and  x  as  the  constant,  to  make 
a  =  0  in  the  successive  steps  of  the  operation.  As  we  would  obviously 
obtain  the  same  result,  with  the  exception  of  an  interchange  of  x  and  a, 
we  may  at  once  write 

,           VI  (m  —  1)  .t'"~^  a^ 
(^x  4-  <^'T  =  •^'"  +  '^^^       «  H ■ — I — 2 — ~~~  "^ 

TO  (to  —  1)  (1^1  —  2)  X™-'  a'  TO  (?»  —  1)  . . .  (m  —  n  +  2)  cc°'-°+'  a"-' 

1.2.3  "^■■■"  1  .2.3.  .  .  (n— 1) 


We  will  repeat  the  laws  in  regard  to  the  exponents  and  coefficients. 
1.  The  first  letter  of  the  binomial  appears  as  the  first  term  of  the 


BINOMIAL    FORMULA.  359 

development,  affected  with  an  exponent  equal  to  tliat  of  the  binomial, 
and  the  exponents  of  this  letter  in  the  successive  terms  decrease  by 
unity  unto  the  last  term,  where  its  exponent  is  zero.  The  exponent  of 
the  second  letter  of  the  binomial  is  zero  in  the  first  teroLi  of  the  develop- 
ment, and  one  greater  in  each  of  the  successive  terms,  and,  in  the  last 
terra  is  m,  the  exponent  of  the  binomial. 

2.  The  sum  of  the  exponents  of  x  and  a,  in  every  term  of  the 
development,  is  equal  to  the  exponent  of  the  binomial.  A  simple  in- 
spection of  the  formula  will  show  this. 

3.  The  coefficient  of  the  first  term  of  the  development  is  unity;  that 
of  the  second  term  is  equal  to  the  exponent  of  the  binomial;  that  of  the 
third  term  is  formed  by  multiplying  the  exponent  of  the  first  letter  of 
the  second  term  by  the  coefficient  of  that  letter,  and  dividing  by  one 
less  than  the  number  denoting  the  place  of  this  required  term ;  that  of 
the  Ji*''  term  is  formed  from  the  (/i  —  l)"*  term,  by  multiplying  together 
the  coefficient  and  exponent  of  the  first  letter  of  that  term,  and  divi- 
ding this  product  by  {n  —  1),  that  is,  by  one  less  than  the  number  of 
the  required  term. 

4.  There  will  always  be  «z  +  1  terms  in  the  development.  For,  we 
get  one  term,  a™,  or  of,  without  differentiating,  and,  since  the  exponent 
of  m  is  diminished  by  unity  for  each  differentiation,  it  is  plain  that  m 
derivatives  can  be  taken  of  (a  +  a:)",  and,  since  we  get  a  term  of  the 
development  by  each  derivation,  we  must  have  in  all  (m  -f  1)  terms. 

5.  The  coefficients  at  equal  distances  from  the  extremes  are  equal. 
For  the  developments  of  (a  -\-  xj"  and  (x  +  a)"  must  be  identical, 
differing  only  in  the  order  of  the  terms;  the  first  term  of  the  develop- 
ment of  (a  -j-  a:)"  being  the  last  term  of  that  of  (x  -\-  «)"",  the  second 
from  the  left  of  the  development  of  (a  +  x)"  being  the  second  from 
the  right  of  that  of  {r,  -f  a)"  ,  &c.  Hence,  the  terms  of  the  develop- 
ment of  (a  -f  a;)",  taken  in  reverse  order,  will  constitute  the  direct 
development  of  (x  -f-  «)"•  I*^  is  plain,  then,  that  the  second  term  from 
the  left  must  have  the  same  coefficient  as  the  second  term  from  the 
right,  the  one  being  formed  from  a"  in  the  same  manner  as  the  other 
from  a™,  and  so,  in  like  manner,  all  the  other  coefficients  at  equal  dis- 
tances from  the  extremes  must  be  equal. 

6.  The  sum  of  the  coefficients  of  the  development  of  {x  -f  a)*"  is 
equal  to  the  m^^  power  of  2.     For,  if  we  make  a-  =  1  and  o  :=  1  in 

the  equation,  (x  +  a)™  =  x""  -\-  ^/i.r"""'  a  -\ ^^ z — -^ + 


360  BINOMIAL    FORMULA. 

m  (m  —  )  (m  —  2)  x""'"  a'  ^  j  v,        -n      j 

— ^^ i~o — ^^ ....  a  ,  the  second  nieinber  will  reduce 

to  the  sura  of  tlic  coefficients,  and  the  first  member  will  become  2"",  and  we 

will  have  (1  +  i)"  =  2-  =  1  +  m  +  m  (m  - 1)  +  ^^'^On—l)(^m—2) 

+  &c.  We  make  x  and  a  unity,  in  order  to  reduce  the  terms  in  the 
second  member  to  their  coefficients;  and,  it  is  plain,  that  if  x  and  a  had 
another  value  than  unity,  the  second  member  would  not  be  the  sum 
of  the  coefficients. 

In  accordance  with  the  demonstration,  the  sum  of  the  coefficients  in 
the  development  (x  +  of,  ought  to  be  4 ;  for,  in  this  case,  m  =  2,  and 
(2)""  =  (2)^  =  4.  And,  in  the  development  of  (x  +  a)^,  m  =  3,  and 
(2)""  =  (2)^  =  8,  which  agrees  with  the  fact.  So,  likewise,  in  the 
development  of  (x  -\-  ay,  m  =  4,  and  (2)"  =  (2)''  =  16,  which  also 
agrees  with  the  fact. 


FORMATION  OF  POWERS  BY  THE  RULE. 

375.  Let  it  be  required  to  find  the  development  of  (x  -\-  of  by  the 
rule.  The  first  term  is  x^,  the  second  term  has  5  for  a  coefficient,  and 
the  exponent  of  x  is  one  less  in  this  term  than  in  the  preceding ;  a 
also  enters  to  the  zero  power  in  the  first  term ;  and,  since  the  expo- 
nents of  a  go  on  increasing  by  unity,  a  must  enter  to  the  first  power  in 
the  second  term.  Hence,  the  second  term  is  5x'*a.  The  coefficient  of 
x  in  the  third  term  is  formed  by  multiplying  5,  the  coefficient  of  x  in 
the  second  term,  by  4,  its  exponent,  and  dividing  the  product  by  2, 

5x4 
the  number  of  terms  preceding  the  required  term.     Hence,  —^ —  =10, 

is  the  coefficient  of  the  third  term,  and,  from  the  law  of  the  exponents, 
X  must  enter  to  the  third  power,  and  a  to  the  second  power  in  the 
third  term.     Hence,  that  term  is  lOx^a^     For  the  coefficient  of  x  in 

10     3 

the  fourth  term,  we  have  — ^ —  =  10,  and  the  fourth  term  must  be 

10  .  2 
lOx'^a".     For  the  coefficient  of  the  fifth  term  we  have  — - —  —  5,   and 

that  term  itself  must  be  5xa''.     For  the  coefficient  of  the  sixth  term 

5  (1) 
we  have  — = —  =  1,  and  that  term  itself  must  be  x^a\  or  a'.     For  the 

5 

coefficient  of  the  seventh  term  we  have  — ^  =  0.    'Hence,  there  is  no 


BINOMIAL    FORMULA.  3G1 

seventl>  term.  Then,  the  development  of  {x  +  a)*  is  x^  +  ox*a  + 
lOx'a^  +  lOx^a^  +  5xa*  +  o',  and  we  see  that  the  sum  of  the  coeffi- 
cients is  equal  to  (2)^  =  32,  and  that  the  sum  of  the  exponents  of  x  and 
a  in  each  term  is  5. 

The  preceding  development  might  have  been  obtained  directly  from 

,  ^         ,     ,         .                           ,        m(m  —  l).c'"~V      ,, 
the  general  formula,  (^x+  a)   =  a;"  -f  mx^  'a  -\ — h  &c., 

by  making  m  =  5.  But  the  above  process  is  generally  shorter  when 
m  is  positive  and  entire. 


DEVELOPiMENT  OF  [x  —  a)'". 

376.  The  development  of  (x  —  a)"  may  be  obtained  in  the  same 
way  as  that  of  (x  +  a)".  But  it  may  be  gotten  at  once  from  the  gene- 
ral formula  by  changing  +  a  into  —  a  ;  the  terms  involving  the  odd 
powers  of  a  will  all  be  negative,  and  we  will  have  (j:  —  a)°  =  af  — 

mx'"-ki   +        ^     ^    .^^ ^  ^     ^'    g + ±  a". 

The  last  term  will  be  positive  when  m  is  an  even  number,  and  nega- 
tive when  m  is  an  odd  number. 

It  will  be  seen  that  all  the  laws  in  regard  to  the  development  of 
(x  4-  a)"  hold  good  in  regard  to  the  development  of  (x  —  a)",  except 
that  the  alternate  terms  must  be  affected  with  the  negative  sign,  and 
that  the  sum  of  the  coefficients  is  zero. 

A  few  examples  will  illustrate  the  use  of  the  two  formulas  for  the 
development  of  (.x  -f  «)■"  and  (x  —  a)". 


EXAMPLES. 

1.  Required  the  sixth  power  of  (.x  +  a). 

Ans.  x^  +  Gx'a  -f  Idx'a''  +  20.x='a='  +  15x^a'  -f  Gxa'  +  a^. 

For,  m  =  G,m  —  l  =  5,  (vi  —  2)  =  4,  «i  —  8  =  3,  m  —  4t  =  2, 

„           ,           ^                              ,       m  (in  —  l)^"-^ 
wi  —  0  =  1.       Hence,  {x  +  a)""  =  x'"  +  mx'"-^  -\ ^ -y-- 

-f- ....  a™  becomes,  in  this  case,  {x  +  ay=x^-\-Qx^  -i '■ — a* 


=  x^  +  Qx'  -f  15xW  -f a«. 

2.  Required  the  development  of  (x  +  ay. 
Ans.  cc"  +  7x^a  +  2lx'a^+S6xW+S5xhi*  +  21x^a^+  Ixa^  -f  a?. 
31 


362  BINOMIAL    FORMULA. 

o.  Develop  (.c  +  ay. 

Ans.  .x^  +  8x-^a  +  28.rV+5GxV+70.^V+56*V-f28.i-aH8xa'+a'. 

4.  Develop  (x  —  af. 

Ans.  x^  —  Q,x^a  +  15.xV  —  •ZQxht'  +  15a;V  —  Qxa^  +  a\ 

5.  Develop  (,r  —  o)'. 

Ans.  x'  —  nx^a  +  21.'r5a2  —  35a;V+35:r'a^  —  21a;2rt  +  ^Ixa^  —  a\ 

6.  Required  the  development  of  {x  —  a)". 

Ans.  x^  —  ^x^a  +  36xV  — 84xV  +  126xV  —  126a'V  +  84xV 
—  36a;V  +  9xa«  +  a». 

377.  The  binomial  formula  has  been  deduced  upon  the  hypothesis 
that  the  coefficients  and  exponents  of  x  and  a  were  unity,  but  it  can 
be  applied  to  binomials  in  which  these  conditions  are  not  fulfilled,  by 
substituting  for  the  letters  within  the  parenthesis  other  letters  whose 
coefficients  and  exponents  are  unity.  Thus,  to  get  the  fifth  power  of 
z^  +  Zy",  let  z^  =  X,  and  3y/^  =  a  ;  then,  {z^-  +  ^iff  =  («  +  aj  =  x' 
-\-bx'^a  +  lOcc'a^  +  lOx^a?  +  bxa'^  +  a^,  which,  by  substituting  for  x  and 
a  their  values,  will  be  found  equal  z^°  +  Ibz^f  -\-  OO^y  +  270^y  + 
^{)bzhf  -f  243^'°. 

In  like  manner,  we  may  find  the  development  of  (nx^  -\-  ra^)'"  = 

m  (m  —  1)  r2«"'-=£cP'"--''a2^ 
jr.rP"  +  7?ir?i'"-'.r'""-P«i  +  — ^i ^-—^ 1-  ■  •  ■  r'"a'^"\    (A) 

In  this  general  formula,  make  x  =  1,  and  a  =  1  •  the  second  mem- 
ber will  reduce  to  the  sum  of  the  coefficients  of  x,  and  the  first  member 
will  become  (n  -f  i-)"".  Hence,  the  sum  of  the  coefficients  of  any  bino- 
mial develojnnent  is  equal  to  the  m**  power  of  the  sum  of  the  coefficients 
within  the  2}cirenthesis. 

In  accordance  with  this  general  formula,  the  sum  of  the  coefficients 
of  the  development  of  (2^  +  oy^y  ought  to  be  equal  to  (4)^  =  1024; 
and  this  agrees  with  the  development  above,  for  1  +  15  +  90  +  270 
+  405  +  243  =  1024. 

378.  Formula  (A)  shows,  moreover,  that  if  the  exponents  of  x  and 
a  are  equal,  that  is,  p  —  q,  the  sum  of  the  exponents  of  x  and  a  in 
each  term  of  the  development  will  p?n.  Whenever,  then,  the  exponents 
ofs.  and  a  arc  equal  in  any  binomial,  the  sum  of  the  exp)onents  of  these 
letters  in  each  term  of  the  development  will  be  equal  to  the  common  ex- 
ponent xoithin  the 'parenthesis  into  the  exponent  of  the  poxoer  to  xohich 
the  binomial  is  to  be  raised. 


BINOMIAL     FORMULA.  863 

The  two  important  laws  just  deduced  from  formula  (A)  are  general 
for  all  binomials.  The  second  and  sixth  laws,  observed  to  govern  the 
development  of  (x  +  a)",  are  but  particular  cases  of  the  foregoing. 

The  binomial  formula  can  be  extended  to  polynomials,  by  representing 
the  polynomial  by  the  algebraic  sum  of  two  letters.  Thus,  to  obtain 
the  square  of  cc'  +  4:X^  —  8x  —  8,  represent  x'^  -f  4:X?  by  x,  and  —  Sx 
—  8,  by  a.  Then,  (x»  +  Ax""  —  Sx  —  Sf  =  (x  +  of  =  x"  +  2,ax 
4-  a^  =  (by  replacing  x  and  a  by  their  values),  x®  -f  8x*  —  80x'  + 
l'28.c  +  64.  In  like  manner,  to  obtain  the  cube  of  x^  -\-  2x  —  4, 
let  x"  =  X,  and  2x  —  4  =  a;  then,  (x'  +  2x  —  4)'  =  (x  +  af=  x" 
4-  3x^a  +  3a^x  +  a',  which,  by  replacing  x  and  a  by  their  values, 
becomes  x«  +  6x^  —  40x'  +  96x  —  64. 

GENERAL   EXAMPLES. 

1.  Find  the  coefficient  of  the  twelfth  term  of  the  development  of 
(.X  +  a'y,  by  means  of  the  formula  for  the  n^^  term. 

m(m.  —  \)(m  —  1)(m  —  Z)  (m  —  i)  (m  — 5)  (to  — 6)  (m  — 7)  (m  — 8)  (»i  — 9)  (m  — 10 
^^'  1X2X3X4X6X6X7X8X9X10X11 

2,  Find  the  twelfth  term  of  the  development  of  (x  +  a)^. 
50  X  49  X  48  X  47  X  46  X  45  x  44  x  43  x  42  x  41  x  40a3V' 


Ans. 


1x2x3x4x5x6x7x8x9x10x11 


3.  Find  the  development  of  (x  +  a)^°. 

Ans  x^o  +  20x'»a  +  lOOx'^a^  +  1140x'V  +  4845x'V  +  15504x%^ 
+38760x'V+77520x'V  +  125970x'V  +  167960x"a«+  184756xV 
+  167960xV'  +  125970.rVt'2  +  77520x^a"  +  38760x«a'*  +  15504x^a'^ 
+  4845x''a'«  +  1140x='a'^  +  190x^a'«  +  20xa'«  +  a^". 

4.  Find  the  development  of  (x^  +  2ax  —  Aa?y. 

Ans.  x^  +  Qax^  —  40o'x^  +  96alr  —  64a«. 

5.  Find  the  4th  power  of  Sa^c  —  2hd. 

Ans.  81aV  — 216aVifZ-f-  2lQa\%hP  —  maWiP  +  lU'd\ 

6.  Find  the  cube  of  x^  —  2x  +  1 

Ans.  x«  —  6x5  ^  i^.jA  _  20a;3  +  15x-  —  6x  +  1. 

7.  Required  the  coefficient  of  the  sixth  term  of  the  development 
of  (x  +  ay. 

Ans   7(7-1)  (7-2)  (7-3)  (7-4)      7.x6.5x4x3_ 
1.2.3.4.5  ^1.2.3.4.5 


364  BINOMIAL    FORMULA. 


DEMONSTRATION   OF    THE   BINOMIAL   FORMULA   FOR   ANY 
EXPONENT. 

379.   1.  Let  m  be  supposed  negative  and  entire,  then  (x  +  a)""  = 

— ,  which,  hy  ae- 


{x  +  a)'"        .x"  +  ?7ij"~'a  -f  m(m  —  l)x^~hi'^  + 

tually  performing  the  division,  will  be  found  equal  to  x~"'  —  mx~"^^a  — 
m(  —  m  —  l'),^-'""^       m(  —  m  —  1)   (  —  m  —  2')x-'^-^a'^ 

rr2  1.2.3  .  .  .  ± 

Px~^'"«"',  &c.  Hence,  we  have  the  same  law  of  formation  as  when  the 
exponent  is  positive  and  entire.  We  might  demonstrate  this  truth  more 
rigorously  by  assuming  (x  +  a)"""  =  A  +  A'a  +  A"a^  +  A."'a^,  &.C., 
and  proceeding  just  as  we  did  when  m  was  positive  and  entire,  regard- 
ing a  as  the  variable.  But  it  is  not  necessary  to  repeat  the  operation. 
We  have  a  right  to  assume  the  exponents  of  a  to  be  positive  and  entire 
in  all  the  terms  of  the  development;  because,  if  we  expand  any  frac- 
tion whose  numerator  is  a  constant,  and  the  first  term  of  whose  deno- 
minator is  a  constant,  all  the  exponents  of  the  variable  will  be  positive 
and  entire  in  the  development. 

2.  When  m  is  positive  and  fractionah     Let  m  =  — ,  and  assume 

q 

(a  +x)^  =  A  +  A'x  +  A"x'  +  M"x^ A-^r-'CP).  Make  ^  =  0, 

p_ 
and  we  have  A  =  as..     Taking  the  derivatives  of  both  members  of 

(P),  and  dividing  by  dx,  we  have  —(a  +  x')'i  =  A'  -f  2A"a;  -f 
3A"V  -V  &c.  (2).     Making  a;  =  0,  we  get  A'  =  ^a~'-     From  (2), 


there  results  U^  —  l\  {a  -f  3;)Z~'=  2 A"  -f  2  .  ^A"'x  +  &c.  (R). 
q\q  J 

Making  x  =  0,  we  get  A"  =  — (— —  l)-^^     ^^^^^  (^)'  ^^^^^'^  ^^" 

suits  ^{^  —  l\{^  —  2\  (a  +  .t)^~'  =  2  .  3 A'"  -f  &c.  From  which, 


2  .3 


BINOMIAL    FORMULA.  305 

Continuing  the  process  and  replacing  the  constants,  A,  A',  &c.,  by 

2  j:        «    1_  1 

their   values,    equation    (P)    becomes   (a  -f  .r)  i  =  a  i  +  —a  i      x  -\- 

il(^_l)„^V+ii(ii-l)(^-2U-V ±PaT-»-':c-V 

q^q         '  q\q  /  \q  / 

rr2  1.2.3 


DEVELOPMENT  OF    BINOMIALS    AFFECTED  WITH    NEGATIVE 
AND  FRACTIONAL  EXPONENTS. 

380.  When  m,  the  exponent  of  the  parenthesis,  (x  -\-  dy°,  is  posi- 
tive and  entire,  it  is  evident  that  successive  differentiation  will  even- 
tually reduce  that  exponent  to  zero,  and  then  the  next  differential  will 
be  zero.  And,  since  the  constants,  A,  A',  &c.,  of  the  assumed  develop- 
ment, have  been  determined  by  the  successive  differentiation  of 
(x  +  ay,  it  is  plain  that  the  development  will  terminate.  But,  when 
the  exponent  of  the  parenthesis  is  negative  or  fractional,  successive 
differentiation  can  never  reduce  it  to  zero,  and,  therefore,  an  infinite 
number  of  constants  can  be  determined  in  the  equation,  (j:  -\-  a)~'",  or, 

p. 
{x  +  a)^.  .  =  A+  A'x  +  A"x'  +  A'"x'  +  &c.     The  series  will  then 

never  terminate. 

EXAMPLES. 


1.  Develop  :j =  0-  —  ^)~'  ^°*-*'  ^  series. 

Ans.   1  -^  X  +  x^  +  x'^  +  x*  -\-  x^  -{■  &c. 
Make  m  =  — 1,  1  =  a",  and  — x  =  a  in  the  formula  (x  +  a)~°  = 


1  .  2 

2.  Develop  = into  a  series. 

^  1  +  a- 

Ans.  1  —  X  +  X-  —  x^  +  x*  —  x^  -\-&c. 

1  _j_  7^  -L  a;2 

3.  Develop  — :, ^  =  (1  -\-  7 x  -\-  x^  (1  -{-  a;)-'  into  a  series. 

I  -\-  X 

Ans.  1  +  6x  —  5x^  +  5x^  —  5x'  -f  5x'  —  &c. 


4.  Develop into  a  series. 

a  -{-  X 

A  1  X      ,    X'  X^         X*  ^    ,     0 

a        a^        o?       a*        a* 
31* 


366  BINOMIAL    FORMULA. 

6.  Develop  .y/l  +  X  into  a  scries. 


Ans.  =t 


a;       X-   .    x^ 


/-,     ,      X  X-         X"  p      V 


6.  Develop  ^x  +  1  into  a  series. 

Ans.  ±(   /—.   — ____^^ zzz  — &c.y 

7.  Develop  ^/2  =  ^1  +  1  into  a  series. 

Ms.  ±(l+i_i  +Jg_-5^  +  &c.). 


8.  Develop  ^/5  =  v/4  +  1  into  a  series. 

Ans.  -t(2  +  i  — J^+rh  — &c.). 

9.  Develop  (^Ji*  +  ?i'')''  into  a  series. 

^?is.   ±  (m  +  I  m-V  —  gS^m-'ji^  +  _|gm-"w'^—  &c.). 

10.  Develop  (a  +  x)*  into  a  series. 

11.  Develop  (m^  +  rr")    into  a  series. 


12.  Develop  (1  +  cc^)^  into  a  series. 

,         01?        x^       5x®         ., 

^'''-  i  +  T-y+sT-'-^'^- 

13.  Develop  -■  into  a  series. 

^/7l2  +  x" 

^ns.   ±  -(1  -  ^,  +  g-,  -  le^^e  +  728^3  -  &c.  I 

14.  Develop  4/a  +  a;  into  a  series. 

,_      1        X  2       cc''  6        :e^  21       a;"         ■. 

15.  Develop  4/1  +  ^=  into  a  series. 

,        ,       la;       2x^       6a;'       21x*    ,     . 
^'''-  ^+-5-25+125-625   +*'• 


BINOMIAL    FORMULA.  367 


16.  Develop  v/1  —  ^  ^^^^  ^  series. 

17.  Develop  1/2  into  a  series. 

Ans.  1  +  I-  ^+  -,&^-  ^-,  +  &c. 


CONSEQUENCES  OF  THE  BINOMIAL  FORMULA  — SQUARE  OF 
ANY  POLYNOMIAL. 

381.  We  have  seen  that  (a-{-by=a''  +  2ab  +  b%  that  is,  the 
square  of  a  binomial,  is  equal  to  the  sum  of  the  squares  of  its  two  terms, 
plus  the  double  product  of  those  terms.  To  find  the  square  of  a  tri- 
nomial ;  a  +  b  -\-  c,  let  a  +  b  =  s.  Then,  (a  +  i  +  c)^  =  (s  +  c)*  = 
s2  +  2sc  +  c'^^a  +  by  +2c(a  +  b)  +  (?  =  ct"  +  i^  +  c^  +  2ca  -(- 
2cb  +  2ab;  that  is,  the  square  of  a  trinomial  is  equal  to  the  sum  of  the 
squares  of  its  three  terms,  plus  the  double  product  of  these  terms  taken 
two  and  two.  To  find  the  square  of  a  polynomial  of  four  terms,  a  -{-b 
+  c  +  (7,  let  a  +  t  +  c  =  s'  Then,  (ii  ^  b  -\-  c  ■\-  df  =  (s'  +  dj 
=  s'2  +  2.i'il+  rp  =  a'  +  b'  +  c""  +  (P  +  2r/a  +  2db  +  2dc  +  2ca  + 
2cb  -\-  2(ib  ;  that  is,  equal  to  the  sum  of  the  squares  of  all  its  terms,  plus 
the  double  product  of  these  terms  taken  two  and  two.  Now,  it  is  evi- 
dent that  the  same  law  will  hold  good  for  the  square  of  a  polynomial 
composed  of  five  terms.  For  this  square  is  composed  of  the  square  of 
the  first  four  terms  plus  the  square  of  the  fifth  term,  plus  the  double 
product  of  the  fifth  term  by  the  sum  of  the  first  four  terms.  And, 
since  the  square  of  the  first  four  terms  will  give  the  sum  of  the  squares 
of  all  these  terms,  plus  the  double  product  of  these  terms,  taken  two 
and  two,  it  is  evident  that,  when  the  double  product  of  the  fifth  term 
by  the  sum  of  the  other  four,  is  added  to  the  other  double  products, 
we  will  have  the  double  product  of  all  the  terms,  taken  two  and  two ; 
and  it  is  plain,  moreover,  that  when  the  square  of  the  fifth  term  is 
added  to  the  sum  of  the  squares  of  the  other  four  terms,  that  we  will 
have  the  sum  of  the  squares  of  all  the  terms.  The  law  is  then  true  for 
five  terms,  and  it  is  plain  that  the  same  reasoning  can  be  extended  to 
sis,  seven,  and  any  number  of  terms.  Hence,  we  conclude,  that  the 
square  of  any  polynomial  is  equal  to  the  sum  of  the  squares  of  all  the 
terms,  plus  the  double  product  of  all  the  terms,  taken  two  and  tico. 


368  BINOMIAL    FORMULA. 


1.  Kequired  the  square  oi  a  -\-  b  +  c  -\-  d  -i-  e. 

Ans.  a'  +  h"^  +  c'  +  cl?  +  e^  -f  2ah  +  2ac  ^-  2ad  +  2«e  +  2hc  + 
Ihd  +  2he  +  2cd  +  2ce  +  2dc. 

2,  Required  the  square  oi  a  -{-  h  -\-  c  -{•  d  -{-  e  -\-  f. 

Ans.  a^  +  Z.2  +  c^  +  d'  -{-c'  +f^  +  2ah  +  2ac  +  2ad  +  2ae  + 
2rr/+  2hc  +  2M  +  2&e  +  2//+  2cf7  +  2ce+  2c/ +  2c?e+  2d/+2cf. 


CUBE  OF  ANY  POLYNOMIAL. 

382.  The  cube  of  (a  +  &),  is  a?  +  Sa^i  +  3a6^  +  h\ 
To  fiud  the  cube  of  a  +  i  4-  c,  let  a  +  hz=  s.  Then,  (a  +  6  +  cf 
=z  (s  +  c)'  =  s^  +  3s2c  +  3sc^  +  c«  =  a^  +  ?y^  +  c'  +  ^a?h  +  Sa^c  + 
36^c  +  3c^a  +  3c^6  +  Qahc ;  that  is,  the  cube  of  a  trinomial  is  com- 
posed of  the  sum  of  the  cubes  of  its  three  terms,  plus  the  sum  of  the 
products  arising  from  multiplying  three  times  the  square  of  each  term 
into  the  first  power  of  the  other  terms,  plus  six  times  the  product  of 
all  the  terms,  taken  three  and  three. 

By  pursuing  precisely  the  same  course  of  reasoning  that  we  have 
already  employed,  it  can  easily  be  shown  that  the  law  is  general  for  any 
polynomial. 

EXAMPLES. 

1.  Required  the  cube  of  «  +  &  +  c  +  d. 

Ans.  a^  +  P  +  c"  +  d^  +  3a^6  +  ^ah  -f  2,ahl  +  U^a  +  Wc  + 
Wd  +  3c^a  +  3c^6  +  3A?  +  M^a  +  M^h  +  M^c  +  Qahc  +  Qahd  + 

f\nr/7  _L   (\Ti/^i7 


Qacd  +  Qbcd, 


BINOMIAL    FORMULA.  bOU 

EXTRACTION  OF   THE  Xth  ROOT  OF  WHOLE  NUMBERS  AND 
POLYNOMIALS. 

383.  The  most  important  consequence  of  the  binomial  formula  is, 
that  we  are  enabled  by  it  to  extract  high  roots  of  whole  numbers  and 
polynomials.  We  will  begin  with  the  simplest  case,  the  extraction  of 
the  «"  root  of  whole  numbers. 


EXTRACTION  OF  THE  Xih  ROOT  OF  WHOLE   NUMBERS. 

Let  a  represent  the  tens,  and  b  the  units  of  the  required  root.  Then, 
the  given  number  will  be  (a  +  />)",  and  from  the  binomial  formula  wo 
have 

(a  +  iy  =  a-  +  na--^b+     ^     ^     ^ +  ^ tV^S^ +  '-^'''- 

That  is,  a  number,  whose  n^^  root  is  composed  of  tens  and  units,  is 
made  up  of  the  ??<■'  power  of  the  tens,  plus  n  times  the  (it  —  1)""  power 
of  the  tens  into  the  first  power  of  the  units,  plus,  &c. 

It  is  evident  that  the  Jt"  power  of  the  tens  will  give,  at  least,  ?*  +  1 
figures,  therefore,  the  »*■»  root  of  the  tens  cannot  be  sought  in  the  7i 
right  hand  figures.  These  must,  therefore,  be  cut  off  from  the  right. 
We  next  seek  the  greatest  n^^  root  contained  in  the  left  hand  period, 
and,  when  we  have  found  it,  we  will  have  a  of  the  formula.  Raise  the 
root  found  to  the  Ji*'  power,  and  subtract  this  power  from  the  left  hand 

period.     The  remainder  will  correspond  to  na^~^h  +   —  -^— ^~^ + 

kc,  of  the  formula.     The  appi-oximate  divisor  to  find  h  (the  unit  of 

xi             x^    •           n   I       mi       X           T    •         •          „   1         "  ("  —  V)n'^^h 
the  root)   is   na""'.     The  true  divisor  is  ;ia°~'  +     -^= — = — -^ + 

n  (n  _  1)  (n  —  2)«"-^i=  ,   ,       ,    ,        ,  •        ,  '  ^'      , 
1~9 — 5 1  ^^  '   "'^^>  ^^  '■'  ^^  unknown,  we  can  only  use 

the  approximate  divisor.  The  n — 1  figures,  on  the  right  of  the  remainder, 
must  be  separated  from  the  other  figures,  because  n  times  the  (n  —  1)"" 
power  of  the  tens  will  give,  at  least,  n  figures.  Dividing  the  period  on 
the  left  by  the  approximate  divisor,  we  get  the  units  of  the  root,  or,  gene- 
rally, a  number  greater  than  the  units,  because  our  divisor  is  too  small. 
It  is  plain  that  the  number  of  ternis  rejected  when  using  the  approximate 
divisor,  depends  upon  the  number  of  units  in  n,  and  that  the  value  of 
each  term  depends  mainly  upon  a.  When,  therefore,  n  and  a  are  both 
large,  or  when  one  only  is  large,  the  approxiinate  divisor  is  very  consider- 
ably too  small,  and  the  quotient,  therefore,  will  be  too  great.    Raise  the 

Y 


370  N      BINOMIAL     FORMULA. 

two  figures  of  the  root  found  to  the  «*'■  power,  and  compare  the  result 
with  the  given  number;  if  it  be  greater  than  this  number,  the  last  figure 
of  the  root  must  be  reduced  by  one,  or  more.  Let  us  illustrate  by  an 
example. 

Required  the  fifth  root  of  33554432. 

885  54432  I  32 
243  I 

«a»-'  =  5  X  81  =  405     92  "54432 

We  began  by  cutting  off  the  five  right  hand  figures.  The  next  step 
was  to  find  3,  the  greatest  fifth  root  contained  in  335,  the  left  hand 
period.  Next,  the  approximate  divisor,  405,  was  formed,  and  this  was 
found  to  enter  twice  in  the  left  hand  period  of  the  remainder,  after 
subtracting  the  n^^  power  of  the  tens  from  the  given  number.  Upon 
trial,  32  is  found  to  be  the  true  root,  for,  (32)*  =  33554432.  In  this 
case,  the  unit  figure  of  the  root  not  being  large  in  comparison  with  the 
tens,  the  approximate  did  not  differ  so  materially  from  the  true  as  to 
give  a  quotient  figure  too  great  by  unity,  or  some  other  number.  But, 
suppose  it  were  required  to  extract  the  fourth  root  of  230625. 

Then,  39*0625  1  25 

16 


jia"-'  =4  X  {2f  =  32,  230  625 

The  approximate  divisor  gives  7  as  a  quotient ;  but,  upon  trial,  it 
has  (o  be  diminished  by  2;  and  25,  not  27,  is  the  true  root. 

It  will  be  seen  that  the  units  of  the  root  can  only  be  ascertained  by 
trial.  If  the  units  of  the  root  be  large  in  comparison  with  the  tens, 
the  quotient  obtained  by  dividing  the  left  hand  period  of  the  remainder 
by  the  approximate  divisor,  will  often  differ  considerably  from  the  true 
units  of  the  root.     The  following  example  will  illustrate  this  : 


</ 


13  0321  =  19 

1 


««"-•  r=:  4  .  1'  =  4       120'321  =  mf-'b  +  "  ^"       \^""  '^''  +  &c. 

In  this  example,  the  approximate  divisor  gives  80  as  a  quotient,  but 
this  is  absurd.  We  try  9,  the  greatest  figure  of  the  units,  and  find  it 
to  be  right;  for  (19)'  =  130321.  The  quotient,  30,  being  so  large,  we 
concluded  that  the  units  of  the  root  must  be  large,  and,  therefore, 
tried  9. 


BINOMIAL    FORMULA.  ^^Tl 

Our  reasoning  has  been  confined  to  numbers  whose  root  contained 
but  two  figures ;  but,  as  in  the  corresponding  cases  of  the  extraction 
of  the  square  root  and  cube  root  of  whole  numbers,  it  can  readily  be 
extended  to  numbers  whose  root  contains  any  number  of  figures.  It  is 
not  necessary  to  repeat  the  generalization  of  the  principles,  and  we, 
therefore,  pass  at  once  to  the  rule  for  the  extraction  of  the  n""  root  of 
whole  numbers. 

RULE. 

I.  Divide  the  given  number  into  periods  of  n  figures  each,  begin- 
ning on  the  right.  Extract  the  n""  root  of  the  greatest  n"*  jMwer  con- 
tained in  the  left  hand  period,  and  set  this  root  op,  the  right,  after  the 
manner  of  a  quotient  in  division.  Subtract  the  n'^  poicer  of  the  root 
so  found  from  the  left  hand  period. 

II.  To  this  remainder  annex  the  next  period,  and  separate  from  the 
new  number  so  formed  the  n  —  1  figures  on  the  right.  Regard  the  left 
hand  period  as  a  dividend. 

III.  Divide  the  dividend  by  n  times  the  (n  —  J^  power  of  the  first 
figure  of  the  root.  The  quotient  will  be  the  second  figure  of  the  root, 
or  a  number  greater  than  the  true  second  figure  of  the  root.  Raise  the 
two  figures  of  the  root  found  to  the  n"'^j02oe?-.  If  the  residt  exceed  the 
two  left  hand  periods  of  the  given  member,  the  last  figure  of  the  root 
must  be  reduced  %intil  the  v^'^  power  of  the  two  figures  is  equal  to,  or 
something  less  than  the  ttco  left  hand  periods. 

IV.  Annex  the  third  period  to  the  remainder^  after  subtracting  the 
vl^ power  of  the  first  two  figures  of  the  root  from  the  two  left  hand 
periods,  and  cut  off  n  —  1  figures  from  the  right  of  the  new  number 
thus  formed,  and  regard  the  left  hand  period  as  a  new  dividend. 

V.  Divide  this  dividend  by  n  times  the  (a  —  1)""  poioer  of  the  first 
tioo  figures  of  the  root,  the  quotient  will  be  the  third  figure  of  the  root, 
or  a  number  greater  than  the  third  figure.  Ascertain,  by  trial,  whether 
the  last  figure  is  correct,  and  proceed  in  this  way  until  all  the  periods 
are  brought  down.  The  number  of  figures  in  the  root  will  ahcays  be 
equal  to  the  number  of  periods  in  the  given  number. 


1.  Required  the  </68574961.  Ans.  91. 


2.  Required  the  ^6240321451.  Ans.  91. 


o72  BINOMIAL    FORMULA. 

3.  Required  the  ^2476099.  Ans.  19 


4.  Required  tlie  V 1082432 16.  Ans.  101 


5.  Required  the  ^1(592662195786551.  Ans.  1111 


6.  Required  the  ^539218609632.  Ans.  222 


7.  Required  the  ^35831808.  Ans.  12 


I  Required  the  ^587068342272.  Ans.  48 


9.  Required  the  V 75 1547478 10816.  A7is.  96 


10.  Required  the  ^54165190265169632.  Ans.  2222 

APPROXIMATE   ROOT   OF  AN   IRRATIONAL  NUMBER   TO 
WITHIN  A   CERTAIN  VULGAR  FRACTION. 

384.  The  principles  of  approximation  have  been  so  fully  explained 
under  the  head  of  the  Square  Root  and  Cube  Root,  that  it  will  only  be 
necessary  now  to  give  the  rule  for  the  approximate  ri*''  root  to  within  a 
vulgar  fraction,  whose  numerator  is  unity. 

RULE. 

Multiply  and  divide  the  given  number  hy  the  n*'»  power  of  the  deno- 
minator of  the  fraction  that  marks  the  degree  of  approximation.  The 
root  of  the  numerator  of  the  new  fraction  thus  formed,  to  within  the 
nearest  unity  divided  hy  the  exact  root  of  the  denominator,  will  he  the 
apjyroximate  root  required. 

EXAMPLES. 

1.  Required  V6  to  within  2.  Ans.  4- 


2.  Required  ^9  to  within  *.  Ans.  #. 


8.  Required  ^90  to  within  }.  A 


ns. 


7 


4.  Required  ^2098152  to  within  i.  Ans.   \^,  or  8,  nearly. 


5.  Required  ;y 8589929092  to  within  f      Ans.  %\  or  8,  nearly. 


BINOMIAL    FORMULA.  373 

APPROXIMATE   ROOT  OF  WHOLE   NUMBERS   TO  WITHIN   A 
CERTAIN  DECIMAL. 

RULE. 

385.  Annex  as  mani/ periods  of  ciphers  (each  period  consisting  o/n 
figures,)  as  there  are  places  required  in  the  root.  Extract  the  n"*  root 
of  the  new  numher  thus  formed  to  within  the  nearest  unit,  and  point 
from  the  right  for  decimals,  the  numher  of  places  required  in  the  root. 


1.  Required  ^39  to  within  .1.  Ans.  2.1 


2.  Required  ^116857201  to  within  .1.  Ans.  41.1. 


3.  Required  V999«^J89  to  within  .01.  Ans.  9.99. 


4.  Required  i/258991  to  within  .1.  Ans.  12.1. 


5.  Required  ^511-999999999999999888  to  within  .01. 

Ans.  1.99. 

APPROXIMATE   Nth  ROOT  OF  A  MIXED   NUMBER  TO  AVITIIIN 
A   CERTAIN   DECIMAL. 

RULE. 

386.  Annex  ciphers,  if  necessary,  xintil  the  decimal  part  will  contain 
as  mani/ periods  of  n  figures  each  as  there  arep)lac.es  required  in  the  root. 
Extract  the  n"*  root,  and  point  from  the  right  the  required  numher  of 
decimal  places. 

EXAMPLES. 

1.  Required  '^1.0051  to  within  .1.  Ans.  1.1,  nearly. 

2.  Required  ^2.488Tto  within  .1.  Ans.  1.2. 


3.  Required  ^3. 583181  to  within  .1.  Ans.  1.2. 


4.  Required  ^62403.21461  to  within  .1.  Ans.  9.1. 


5.  Required  ^392.5430946993  to  within  .01.  .4ns.  3.31. 


6.  Required  ^1725.4995508234  to  within  .01.  Ans.  4.44. 

32 


374  l>KilMUTATIONS    AND    COMBINATIONS. 

APPROXIMATE  Nth  ROOT  OF  NUMBERS  ENTIRELY  DECIMAL. 

RULE. 

387.  Annex  ciphers,  if  necessary,  until  there  arc  as  many  p>ei'iods  of 
n  figures  each  as  there  are  places  required  in  the  rout.  Extract  the  n"* 
root  of  the  11  ew  number  thus  formed,  and  point  off  the  required  numher 
of  decimal  places. 

EXAMPLES. 


1.  Eequired  ^.00:^45  to  within  .1.  Ans.  .3. 


2.  Required  ^.0028629161  to  within  .01.  Ans.  .31. 


3.  Required  V-0004084201  to  within  .01.  Ans.  .21. 


4.  Required  ;/. 51676701935  to  within  .01.  Ans.  .91. 


5.  Required  y. 96059690  to  within  .01.  Ans.  .99. 


6.  Required  ^.1245732577  to  within  .01.  Ans.  .66. 


^  Required  ^  .464404086785  to  within  .01.  Ans. 


8.  Required  V/.0000000000285311070612  to  within  .01. 

Ans.   .11. 


9.   Required  1^.00048828126  to  within  .1.  Ans.  .5. 


10.  Required  '.y. 600000000000000000000285311670612  to  with- 
in .001.  Ans.  .011. 


11.  Required  y. 2706784158  to  within  .01.  Ans.  .77. 


PERMUTATIONS  AND  COMBINATIONS. 

388.  Permutations  are  the  results  obtained  by  writing  n  letters  lu 
sets  of  1,  2,  3,  ...  .  or  n  letters,  so  that  each  set  shall  difter  from  all 
the  other  sets  in  the  order  in  which  the  letters  are  taken.  Thus,  the 
permutations  of  the  three  letters,  a,  h,  and  c,  taken  singly,  are  a,  h,  c  ; 
and  in  sets  of  two  letters  each,  ah,  ac,  ha,  he,  ca,  cb ;  and  in  sets  of 
three  letters  each,  abc,  ach,  hac,  hca,  cha,  cab. 


PERMUTATIONS    AND    COMBINATIONS.  875 

It  will  be  seen  that  the  sets  differ  only  in  the  manner  in  which  the 
letters  are  written,  the  letters  in  all  the  sets  being  the  same. 

389.  Let  it  he  required  to  determine  the  number  of  permutations  of 
m  letters,  taken  n  in  a  set. 

Suppose  the  letters  to  be  a,  h,  c,  d,  &c.,  and  let  us  first  permute  them 
singly,  we  will  evidently  have  m  permutations  of  the  m  letters  taken 
one  in  a  set.  Now,  let  us  reserve  a,  there  will  remain  m  —  1,  letters 
b,  c,  d,  &c.,  which,  permuted  singly,  will  give  (m  —  1)  permutations,  b, 
r,  d,  e,  f,  &c.  :  write  a  before  each  of  these  letters,  and  we  will  have 
in  —  1  permutations,  ah,  ac,  ad,  &c.,  of  two  letters,  with  a  as  the  first 
letter  of  each  set.  Next,  let  us  resei-ve  b  out  of  the  letters,  a,  b,  c,  d, 
&e.,  and  then  permute  a,  c,  d,  &c.,  singly.  We  will  again  have  (m  —  1) 
permutations  of  the  letters,  taken  in  sets  of  one  letter  each.  Writing  b 
before  each  of  these  sets,  we  will  have  (m  —  1)  permutations  of  n». 
letters,  taken  in  sets  of  two,  with  b  as  the  first  letter.  Reserving  c  in 
like  manner,  we  can  again  form  (m  —  1)  permutations  of  m  letters, 
taken  two  in  a  set,  with  c  as  the  first  letter  of  each  set.  Reserving  all 
the  m  letters  in  succession,  we  can  evidently  form  m  (jm  —  1)  permu- 
tations of  m  letters,  taken  2  in  a  set.  And,  of  these,  a  will  be  the  first 
letter  of  (m  —  1)  sets,  b  the  first  letter  of  (nt  —  1)  sets,  c  the  first 
letter  of  (in  —  1)  sets,  and  so  on.  It  is  plain,  moreover,  that  a  will 
not  only  be  the  first  letter  of  (»i  —  1)  sets,  but  that  it  will  also  be  the 
last  letter  of  (m  —  1)  sets,  and  that  it  therefore  has  been  made  to 
occupy  all  possible  positions.  As  the  same  remark  may  be  made  of  b 
and  all  the  other  letters,  it  is  obvious  that  m  (ni  —  1)  truly  expresses 
the  number  of  permutations  that  can  be  made  of  m  letters,  taken  two 
in  a  set.  That  is,  the  number  of  permutations  of  m  letters,  taken  two 
in  a  set,  is  equal  to  the  total  number  of  letters  into  the  total  number  of 
letters,  less  one. 

In  like  manner,  we  may  find  the  number  of  permutations  of  m  letters, 
taken  three  in  a  set.  For,  if  we  omit  one  of  the  letters,  as  a,  there 
will  be  m  —  1  letters  left,  and  these,  permuted  in  sets  of  two  letters, 
will  give(w — 1)  (m  —  2)  permutations.  For,  we  have  just  seen  that 
the  number  of  permutations,  taken  in  sets  of  two,  was  expressed  by  the 
number  of  letters  into  the  number  of  letters,  less  one.  Writing  the 
reserved  letter,  a,  before  each  of  these  (in  —  1)  (m  —  2)  sets,  we  will 
form  (m  —  1)  (in  —  2)  permutations  of  m  letters,  taken  three  in  a  set, 
with  a  as  the  first  letter  of  each  set.     Reservino;  in  succession  each  of 


370  PERMUTATIONS    AND    COMBINATIONS. 

the  tn  letters,  wc  can  plainly  form  m  {in  —  1)  (in  —  2)  permutations 
of  m  letters,  taken  three  in  a  set. 

By  the  same  course  of  reasoning  it  can  be  readily  shown  that  the 
number  of  peirmutations  of  m  letters,  taken  four  in  a  set,  will  be  ex- 
pressed by  m  (m  —  1)  {in  —  2)  (m  —  3)  ;  and  of  m  letters,  taken  five 
in  a  set,  by  m  (m  —  1)  {m  —  2)  (w  —  3)  (m  —  4). 

We  observe,  in  all  these  expressions,  that  the  first  factor  is  the  number 
of  letters,  and  that  the  last  factor  is  the  number  of  letters,  diminished 
by  the  number  in  a  set  less  one.  Moreover,  each  factor  after  the  first 
is  one  less  than  the  preceding  factor.  Hence,  the  number  of  permu- 
tations of  m  letters,  taken  n  together,  will  be  expressed  by  m  (in  —  1) 
{m  —  2)  (m  —  3) (in  —  n  +  1),  or,  denoting  by  A  the  num- 
ber of  permutations  of  m  letters,  taken  n  in  a  set,  wc  have  A  = 
in  (in  —  1)  (in  —  2) (m  —  n  +  1). 

To  apply  this  formula,  it  is  best  to  determine  in  the  first  place  the 
last  factor,  we  then  know  where  to  stop.  Since  the  first  factor  is 
always  the  number  of  letters,  and  the  law  respecting  the  mean  factors 
is  known,  the  formula  can  then  be  readily  applied. 


1.  Required  the  number  of  permutations  of  10  letters,  taken  7  and  7. 
Ans.  10  (10  —  1)  (10  —  2)  (10  —  3)  (10  —  4)  (10  —  5)  (10  —  6), 

Orl0x9x8x7x6x5x4  =  604800. 

For,  11  =  7,  and  in  =  10,  then  in  — n  -j- 1  =  10 —  7  +  1  =  10 — 
6,  the  last  factor  is  then  4,  and  the  first  10,  the  intermediate  factors  can 
easily  be  formed. 

2.  Required  the  number  of  permutations  of  5  letters,  taken  4  and  4. 

Ans.  5  (5  —  1)  (5  —  2)  (5  —  3)  =  5  X  4  X  3x2  =  120. 

3.  Required  the  number  of  permutations  of  12  letters,  taken  10  and  10, 
Ans.  12  (12  —  1)  (12  —  2)  (12  —  3)  (12  —  4)  (12—5)  (12  —  6) 

(12  —  7)  (12  —  8)  (12  —  9). 

4.  Required  the  number  of  permutations  of  12  letters,  taken  11 
and  11.  Ans.  12x11x10x9x8x7x6x5x4x3x2. 

5.  Required  the  number  of  permutations  of  12  letters,  taken  12  in 
a  set. 

Ans.  12  (12  —  1)  (12  —  2)  (12  —  3)  (12  —  4)  (12  —  5)  (12  —  6) 
(•12  —  7X(12  —  8)  (12  —  9)  (12  —  10)  (12  —  11)  ;  or,  reversing  the 
factors,  1x2x3x4  x5x6x7x8x9x  10  X  11 X 12. 


PERMUTATIONS    AND    COMBINATION; 


Remarks. 

390.  If,  in  the  general  formula,  A  =  m(m  —  1)  (m  —  2)  .  .  .  . 
(m  —  n  -\-  1),  we  make  m  =  n,  and  call  B  the  corresponding  value  of 
A,  we  will  have  B  =  n  (?i  —  1)  (n  —  2)  ....  4  x  3  x  2  X  1 ;  or, 

reversing  the  order  of  the  factors,  B  =  lx2x3x4 {n  —  2) 

(n  —  V)n.  Hence,  the  number  of  permutations  of  n  letters,  taken  all 
together,  is  equal  to  the  product  of  all  the  natural  numbers  from  1  to 
n,  inclusive. 

Example  5  is  an  illustration  of  this. 

It  is  plain  that,  by  increasing  n  (jti  remaining  constant),  we  will  in- 
crease the  number  of  permutations,  because  we  will  have  more  factors 
in  the  product  that  expresses  the  number  of  permutations.  And,  when 
n  is  made  equal  to  m,  this  product  is  the  greatest  possible.  Thus,  the 
result  is  greater  in  example  4  than  in  3,  and  greater  in  5  than  in  4. 

When  n  =  m  +  1,  the  whole  formula  reduces  to  zero.  This  plainly 
ought  to  be  so,  since  it  is  impossible  to  permute  m  letters  in  sets  of 
m+  1  letters.  Zero,  here,  Ss  in  many  other  places,  is  the  symbol  of 
impossibility.  • 

When  n  =  1,  the  last  factor,  m  —  n  +  1,  is  equal  to  m.     Hence, 

the  first  and  last  factors  are  the  same,  and  we  have  A  =  ^-  =  m  ;  that 

is,  the  number  of  permutations  of  m  letters,  taken  one  in  a  set,  is  equal 
to  m. 

COMBINATIONS. 

391.  Combinations  are  the  results  obtained  by  writing  any  number 
of  letters,  as  m,  in  sets  of  1  and  1,  2  and  2,  3  and  3,  ....  w  and  n, 
m  and  m,  so  that  each  set  shall  differ  from  all  the  other  sets  by  at  least 
one  letter.  In  permutations,  the  sets  differ  in  the  order  in  which  the 
letters  are  written ;  in  combinations,  the  sets  differ  in  the  letters  them- 
selves. Thus,  the  letters,  abc,  taken  all  together,  give  but  one  combi- 
nation, abc;  but  will  give  six  permutations, 

abc,  acb,  bar,  bca,  cab,  cba. 

If  the  same  letters  are  taken  two  and  two,  they  will  give  but  three 
combinations 

ab,  ac,  be, 

while  each  combination  is  susceptible  of  two  permutations ;  and  there 
32* 


378 


PERMUTATIONS  AND  COMBINATIONS, 


are,  therefore,  six  permutations  of  the  three  letters,  taken  two  aud  two, 
thus, 

ab,  .„     .       }  cc,        ^  ^       .^^     .        I  he. 


1  "^^}  ) 

ah  will  give    f  ,       ac  will  give    > 


and  he  will  irive     ,    , 
ca,  '^         J  ch. 


392.  Let  it  he  o-equircd  to  determine  the  total  immhcr  of  comhtna- 
nations  of  m  letters,  tahen  n  in  a  set. 

It  is  evideut,  from  what  has  been  shown,  that  if  we  knew  the  total 
number  of  combinations  of  m  letters,  taken  n  in  a  set,  that  we  could 
get  the  number  of  permutations  of  m  letters  taken  n  in  a  set,  by  mul- 
tiplying the  number  of  combinations  by  the  number  of  permutations 
obtained  by  permuting  each  combination ;  or,  in  other  words,  by  multi- 
plying the  number  of  combinations  of  m  letters,  taken  n  in  a  set,  by  the 
number  of  permutations  of  n  letters,  taken  all  together.  Conversely, 
when  the  number  of  permutations  of  m  letters,  taken  n  in  a  set,  and 
the  number  of  permutations  of  n  letters,  taken  all  together,  are  known, 
we  can,  by  dividing  the  former  by  the  latter,  determine  the  number  of 
combinations  of  m  letters,  taken  n  in  a  set. 

393.  To  illustrate  more  fully  by  an  example. 

Let  us*  combine  the  four  letters,  a,  h,  c  and  d,  in  sets  of  three,  we 
will  have  the  four  combinations, 

ahcy  ahd,  acd,  hcd. 

If,  now,  we  permute  each  of  these  sots,  taking  all  the  letters  in  each 
set,  we  will  have  the  total  number  of  permutations  of  four  letters,  taken 
three  in  a  set,  that  is,  the  total  number  of  permutations  of  vi  letters, 
taken  n  in  a  set. 

"Writing  the  results  in  tabular  form,  we  have 


.    ahc 

ahd 

acd 

hcd 

r 

ahe 

ahd 

acd 

hcd 

ach 

had 

adc 

hdn 

Permutations  of  each  combination,  taken  3 

hac 

Ida 

cad 

eld 

and  3,  or  all  together. 

hca 

dah 

cda 

cdh 

cah 

dba 

dac 

dbc 

cha 

adh 

dca 

dch 

We  see  that  each  of  the  four  combinations,  in  sets  of  3,  gives  six 
permutations,  taken  3  and  3.  There  will,  therefore,  be  4  X  6  =  24 
permutations  of  four  letters,  taken  three  in  a  set.  Now,  as  a  corres- 
ponding table  could  be  formed  for  any  number  of  letters,  it  is  plain  that 


PERMUTATIONS    AND    COMBINATIONS.  379 

the  total  number  of  permutations  of  m  letters,  taken  n  in  a  set,  is  equal 
to  the  number  of  combinations  of  m  letters,  taken  n  in  a  set,  multi- 
plied by  the  number  of  permutations  of  n  letters,  taken  all  together. 

Hence,  if 

X  =  the  number  of  combinations  of  m  letters,  taken  n  in  a  set ; 
Y  =  the  number  of  permutations  of  n  letters,  taken  all  together ; 
Z  =  the  total  number  of  permutations  of  m  letters,  taken  n  in  a  set, 

we  shall  have  Z  =  X  .  Y.     Hence,  X  ^  y. 

But  Z  and  Y  are  already  known. 

For  Z  =  m  (m  —  1)  (m  —  2) {m  —  ?i  +  1),  Formula  (A)  ; 

and  Yr=  1  .  2  .  3  .  4  .  5 n,  Formula  (B).    Hence,  we  have  X== 

m(m  —  1)  .  .  .  .  (m  —  n  -f  1)       ..^  ...  i   j     xi    i.  xi 

— ^^ — - — r^^ .     Irom  which,  we  conclude  that  the 

1.2.3.4....» 

number  of  combinations  of  m  letters,  taken  n  in  a  set,  is  equal  to  the 

number  of  permutations  of  m  letters,  taken  n  in  a  set,  divided  by  the 

number  of  permutations  of  n  letters,  taken  «  in  a  set. 

In  the  application  of  this  formula,  it  will  be  of  service  to  remember 
that  the  first  factor  of  the  numerator  is  the  number  of  letters,  that  each 
successive  factor  is  one  less  than  the  preceding,  and  that  the  last  factor 
is  the  number  of  letters  diminished  by  the  number  in  a  set,  less  one. 
It  is  well,  then,  to  determine  first  the  extreme  factors  of  the  numerator; 
the  mean  factors  can  then  be  readily  supplied. 

In  regard  to  the  denominator,  the  last  factor  is  the  number  in  a  set, 
and  the  factors  counted  to  the  left  go  on  diminishing  by  unit}'^  unto 
the  first  factor,  which  is  always  unity. 


1.  Required  the  number  of  combinations  of  6  letters,  taken  4  in  a  set. 
6(6  — 1)  (0  —  2)  (6  — 3)       G.x5x4x3       ^^ 
1 . 2 . o . 4  1.2.3.4 

For  m  =  6  and  n  =  4.  Hence,  the  first  factor  of  the  numerator  is 
6,  and  the  last  m  —  n  ■{■  1  =  6  —  4  +  1  =  6  —  3.  The  two  mean 
factors  of  the  numerator  are  then  readily  formed,  the  extremes  being 
known. 


380  PKRMUTATIONS    AND    COMBINATIONS. 

2.  Required  the  number  of  combinations  of  12  letters,  taken  7  and  7. 
12(12  —  1)  (12  —  2)  (12—3)  (12—4)  (12  —  5)  (12  —  6) 
^'''-  1.2.3.4.5.6.7 

12x11x10x9x8x7x6       12x11x10x9x8 


1x2x3x4x5x6x7  1x2x3x4x5 

X  9  x  8  =  792. 


r^ll 


3.  Required  the  number  of  combinations  of  12  letters,  taken  5  and  5, 
L)(12  — 2)(12— 3)(12— 4)_12xllxl0x 
1x2x3x4x5  ~"  12x10 


12(12  —  1)  (12  — 2)  (12— 3)  (12— 4)_12xllxl0x9x8 


=  792. 

4.  Required  the  number  of  combinations  of  12  letters,  taken  9  and  9. 

12(12—1)  (12—2)  (12—3)  fl2— 4)  (12—5)  (12—6)  (12—7)  (12— 8)  _ 

'■  1.2x3x4x5X6x7x«Xy  ~" 


1X2X3X4X5X6X7X8X9 


5.  Required  the  number  of  combinations  of  12  letters,  taken  3  and  3 

12  (12-1)  (12 -2)  _  12  X  11  X  10  _ 
Ans.  ^2.3  -  6  -  •'^^- 

6.  Required  the  number  of  combinations  of  10  letters,  taken  9  and  9. 

Ans.  10. 

7.  Required  the  number  of  combinations  of  10  letters,  taken  1  and  1. 

Ans.  10. 

8.  Required  the  number  of  combinations  of  25  letters,  taken  21 
and  21.  Ans.  12650. 

9.  Required  the  number  of  combinations  of  25  letters,  taken  4  and  4. 

Ans.  12650. 

10.  Required  the  number  of  combinations  of  50  letters,  taken  47 
and  47.  Ans.  19600, 

11.  Required  the  number  of  combinations  of  50  letters,  taken  3  and  3. 

Ans.  19600. 

Scholium. 

394.  The  last  ten  examples  show  that  the  same  number  of  letters, 
combined  differently,  may  produce  the  same  number  of  combinations, 
that  is,  the  number  of  combinations  of  m  letters,  taken  p  in  a  set,  may 
be  equal  to  the  number  of  combinations  of  ??i  letters,  taken  g'  in  a  set. 
We  propose  to  show  that  this  will  always  be  so  when  m  =p  -^q.    Thus, 


PERMUTATIONS    AND    COMBINATIONS.  381 

in  example  10,  vi  =  50  and  p  =  47 ;  in  example  11,  m  =  50  and 
q  =  o.  The  results  have  been  the  same  in  those  examples,  because 
m  =  p  +  q. 

To  show  this,  suppose  p^  q.     Then  the  number  of  combinations 
of  m  letters,  taken  p  m  sl  set,  will  be  expressed  thus  : 
^  _  m{m — 1)  ....  (ot — g4l)  (??i — q)  (m — q — 1) (711— p  +  1) 

1 .  2Ts (p-iy^  ■ 

And  for  the  number  of  combinations  of  m  letters,  taken  g'  in  a  set, 
we  will  have  the  formula, 

_  «i  (m.  —  I)  (m  —  2) (m  —  9  +  1) 

^  -  1.2.3   (q-l)q  — 

Hence,  D  =  C  (^- 9)  (^ -9- 1)  ■  •  •  •  (^ -^  +  1) 

(?  +  1)  (?  +  2) p 

Now,  it  is  plain  that  D  will  be  equal  to  C,  when  (m  —  q)  (m  —  q  —  1) 
&c.  =p(p  —  1)  .  .  .  .  (q  +  2)  (5' +  1).  Or,  since  tlie  foctors  iu 
both  members  go  on  decreasing  by  unity,  and  since  the  number  of 
them  is  also  equal,  the  last  equation  will  be  true  when  m  —  q  =:p,  or 
in  =  q  -\-  p>,  fis  enunciated. 

This  rule  is  of  importance  whenever  _?J  ^  -^^  • 

For,  then,  we  have  only  to  take  the  difference  between  /)  and  m,  and 
the  number  of  combinations  of  m  letters,  taken  m  — p  and  m  — p,  will 
be  the  same  as  the  number  of  combinations  of  m  letters,  taken  p  and  p. 
Thus,  in  example  10,  instead  of  taking  the  number  of  combinations  of 
50  letters,  taken  47  in  a  set,  we  may  take  the  number  of  combinations 
of  50  letters,  taken  50 — 47,  or  3  in  a  set. 

In  like  manner,  the  number  of  combinations  of  100  letters,  taken  90 
in  a  set,  would  be  the  same  as  the  number  of  combinations  of  100 
letters,  taken  10  in  a  set. 

395.  If,  in  the  formula,  D  =  C  (^-g)  («^-g-l)  •  •  • -(^^-p  +  D 

(9  +  1)  (9  +  2)  .  .  .  .  p         ' 

we  make  q  =  v  —  1,  we  will  have  D  =  C :    for,   in  that 

case,  J)  will  have  only  one  more  factor  than  C.  Hence,  the  number  of 
covibinations  of  m  letters,  taken  p  and  p,  is  equal  to  the  number  of 
combinations  of  m  letters,  taJcen  p  —  1  and  p — 1,  multiplied  by  the 

factor When  p  =  2,  3,  4,  &c.,  or  the  sets  are  made  up 

of  2,  3,  4,  &c.,  letters,  the  factor ~ will  become  —^ — , 


382  PERMUTATIONS    AND    C  0  Jl  B  I  N  ATI  O  N  S  . 

■ffi 2   m 3 

— ^T — ,  — J — .     Suppose,   for  example,   that  we  have   8   letters,  or 

m  =z  8,  then  the  successive  multipliers  will  be  |,  |,  |,  |,  |,  '^.  And, 
since  the  number  of  combinations  of  8  letters,  taken  1  and  1,  is  8,  we 
will  have  the  number  of  combinations,  taken  2  and  2,  3  and  3,  4  and  4, 
&c.,  expressed  8  X  |  =  28,  28  x  |  =  56,  56  X  |  =  70,  70  x  |  =56, 
56  X  i  =  28,  28  x'  f  =  8. 

This  remarkable  law  has  had  many  important  applications,  one  of  the 
principal  of  which  is,  the  determining  of  the  coefficients  in  the  binomial 
formula. 

It  is  plain  from  what  has  been  shown  that,  if  we  write  in  succession 
the  number  of  combinations  of  m  letters,  taken  1  and  1,  2  and  2,  .  .  .  . 
n  and  n,  we  will  have  a  series  of  numbers  increasing  to  the  middle  term, 
and  then  repeated  in  retrograde  order. 

Thus,  6  letters  combined,  1  and  1,  2  and  2,  3  and  3,  4  and  4,  5  and  5, 
give  the  series  6,  15,  20,  15,  6.  In  like  manner,  7  letters  give  the 
scries,  7,  21,  35,  35,  21,  7.  The  law  of  formation  is  precisely  that 
which  we  have  observed  in  the  binomial  development;  and,  in  fact,  in 
this  development,  the  coefficients,  after  the  second,  are  the  combinations 
of  m  letters,  taken  2  and  2,  3  and  3,  &c.,  .  .  .  .  n  and  n. 

396.  Let  us  now  seek  the  greatest  term  of  the  series  formed  by  com- 
bining m  letters,  1  and  1,  2  and  2,  ....  n  and  n.     The  factors,  which 

m  —  1    TO  —  2            m  —  91  4-  1 
determine  the  successive  terms,  are  — - — ,  — ^^ — ,  .... . 

&c.  Now,  since  these  numerators  go  on  decreasing,  and  the  denomi- 
nators increasing,  it  is  evident  that  the  successive  products  will  go  on 
increasing  until  w  —  n  -\-l  =  n,  and  after  that  will  decrease  and  be 
repeated  in  reverse  order.  Suppose  m  an  odd  number,  then  placing 
m  —  m  -f-  1  =  n,  we  get  n  r=  ^  (m  4-  1).  Now,  if  we  include  unity, 
the  last  term  of  the  series,  there  will  be  m  terms  in  all,  and  in  case  of  m 
being  odd,  the  n*""  term  will  obviously  be  even,  and,  of  course,  have  an 
even  number  of  terms  preceding  and  succeeding  it.     Thus,  in  the  series, 

7,  21,  35,  35,  21,  7,  1 ;  «  =  —-, —  =  4,  will  represent  the  4th  term, 

and  we  see  that  it  has  three  terms  before  and  three  after  it.     It  is  plain, 

moreover,  that  tlie  7i^^  term,  being  formed  from  the  (ji  —  l)*""  term  by 

,  .  ,  .        -     ,          ,171  —  n  -{-  1      ^   .      ,  , 

multiplying  the  latter  by  =1,  is,  also,  equal  to  the  (n  —  ]  )<" 

term. 

Hence,  when  m  is  an  odd  number,  there  will  be  two  equal  central 


PERMUTATIONS    AND    COMBINATIONS.  383 

terms,  the  series  will  go  on  increasing  unto  the  first  of  these,  and 
decreasing  from  the  second  unto  the  last  term,  unity.  Thus,  the  com- 
binations of  5  letters,  taken  1  and  1,  2  and  2,  3  and  3,  4  and  4,  5  and 
5,  give  the  series  5,  10,  10,  5,  1. 

397.  When  m  is  even,  the  factors  increase  until  n  =  J  m.  We  can- 
not, in  this  case,  have  n  z=z  — - — ,  for  that  would  make  n  fractional,  a 

manifest  absurdity.  No  factor,  in  this  case,  is  unity,  and,  of  course, 
there  are  no  equal  central  terms.  The  terms  go  on  increasing  until 
n  =  ^m,  and  are  then  reproduced  in  reverse  order.     It  is  obvious  that 

whenever  n  '^Im,  the  factor        will  become  a  proper  fraction, 

and,  of  course,  the  term  of  the  scries  formed  by  multiplying  the  pre- 
ceding term  by  this  factor,  will  be  less  than  the  preceding.  Ten  letters, 
taken  1  and  1,  2  and  2,  &c.,  give  the  scries,  10,  45,  120,  210,  252, 
210,  120,  45,  10,  1. 

398.  If  we  add  together  the  values  of  C  and  ]),  as  found  in  Art. 
394,  and  at  the  same  time  make  q  =  2)  —  1,  wc  will  have  C  +  I)  = 
C  +  C(m-p-f  l)^C(m  +  l)  ^  m-f  1    ^  ^  X  -  =  /"^  +  h 

p  p  1  p       \     1     / 

(m  \   tm  —  1  V  {jn  — p  +  3)  {in  — p  -f  2) 

T/  V   "3     / p  —  \  p        ■ 

The  second  member  of  this  equation  plainly  denotes  the  number  of 
combinations  of  m  +  1  letters,  taken  p  andpj  whilst  the  first  member 
denotes  the  sum  of  the  combinations  of  m  letters,  taken  ^^  andp,  and 
p  —  1  and  p  —  1.  This  important  relation  enables  us  to  get  the  number 
of  combinations  of  m  -}-  1  letters  from  that  of  m  letters.  Thus,  make 
m  =  4,  p  =  3,  and  p  — 1  =  2,  then  4  letters,  taken  2  and  2,  give  6 
combinations,  and  taken  3  and  3,  give  4,  and  we  find  that  5,  or  m  -f  1 
letters,  taken  p,  or  3  in  a  set,  give  a  number  of  combinations  equal  to 
4  +  6,  or  10. 

399.  Upon  this  principle,  the  following  table  (p.  384)  has  been  con- 
structed. The  first  vertical  column  is  made  up  of  ones ;  the  second  column 
contains  the  natural  numbers  from  1  to  20,  and  expresses  the  number  of 
combinations  of  these  numbers,  taken  1  and  1;  the  other  vertical 
columns  express  the  combinations  of  the  same  numbers,  taken  3  and  3, 
4  and  4,  5  and  5,  &c.  These  vertical  columns  result  from  the  hori- 
zontal, which  are  constructed  according  to  the  principle  demonstrated 
in  Article  398,  observing  that  every  horizontal  column  must  close  with 
unity.     Thus,  the  numbers  in  the  second  horizontal  column  are  1,  2 


384 


P  E  R  I\I  U  T  A  T I  O  N  S     AND     COMBINATIONS. 


and  1,  the  numbers  in  the  next  column  are,  1  +  2  =  8,  2  +  1  =  3 
and  1,  and  express  the  number  of  combinations  of  three  letters,  taken 
1  and  1,  2  and  2,  and  3  and  3.  Now,  prefix  unity,  and  we  have  the 
third  horizontal  row  made  up  of  1,  3,  3  and  1,  and  the  fourth  row  will 
be  made  up  of  1,  3  +  1  ^  4,  3  +  3  =  6,  3  +  1  =  4  and  1,  the  last 
four  numbers  expressing  the  number  of  combinations  of  four  letters, 
taken  1  dnd  1,  2  and  2,  3  and  3,  and  4  and  4.  All  the  other  horizontal 
columns  are  formed  in  the  same  manner. 


r- 

ARITHMETICAL  TRIANGLE  OF  PASCAL. 

1 

2 
3 
4 
5 

1 
1 
3 
6 
10 

4 
10 

1 

5 

1 

0 
7 
8 
9 
10 

15 
21 

28 
36 

45 

20 
35 
56 
84 
120 

15 
35 
70 
126 
210 

6 
21 
56 
126 

252 

1 

28 
84 
210 

1 

8 

36 

120 

1 
9 

45 

1 

10 

1 

n 

55 

165 

330 

462 

462 

330 

165 

65 

11 

12 

66 

220 

495 

792 

924 

792 

495 

220 

66 

13 

78 

286 

715 

1287 

1716 

1716 

1287 

715 

286 

U 

91 

364 

1001 

2002 

3003 

3432 

3003 

2002 

1001 

- 

15 

105 

455 

1365 

3003 

5005 

6435 

6435 

5005 

3003 

16 

120 

560 

1820 

4368 

8008 

11440 

12870 

11440 

8008 

17 

136 

680 

2380 

6188 

12376 

19448 

24310 

24310 

19448 

Il 

18 

153 

816 

3000 

8568 

18564 

31824 

43758 

48620 

43768 

19 

171 

969 

3876 

11628 

27132 

60388 

75582 

92378 

92378 

20 

190 

1140 

4845 

15504 

38760 

77520 

125970 

167960 

184756 

0 

1  and  1 

2  and  2 

3  and  3 

4  and  4 

5  and  5 

6  and  6 

7  and  7 

8  and  8 

9  and  9 

10  and  10 

400.  We  can  now  readily  find  the  entire  sum  of  the  combinations 
formed  by  any  number  of  letters,  taken  in  every  possible  way.  Let  the 
numbers  in  the  ^"'  row  be  expressed  hj  1,  a  h  .  .  .  m,  m  .  .  .  b,  a,  1;   t 

being  supposed  an  odd  number.     Then  their  sum  will  be  2  (o  +  6 

-}-  m  +  1).  The  numbers  in  the  next  row  will  be  expressed  by  1  +  a, 
a  +  i  .  .  .  .  +  .  +  m,  +  m  +  m .  . .  i  +  a,  a  +  1  and  1,  and  their  sum 
will  be  4  (a  +  h  . .  .  +  m  +  1)  +  1.  Hence,  the  sum  of  the  numbers  in 
the  (t  +  1)'"  column,  will  be  double  of  the  preceding,  and  one  more. 
When  t  is  an  even  number,  the  same  law  can  be  shown  to  be  true 
But  the  sum  of  the  numbers  in  the  second  horizontal  column,  is  2^  —  1 
Hence,  in  the  third  column,  it  will  be  2  (2'  —  1)  +  1  =  2*  —  1,  and 
in  the  fourth,  2  (2='—  1)  +  1  =  2^  —  1.  And  it  is  plain  that,  in  the 
f"  column  it  will  be  2*  —  1.     It  will  be  seen  that  the  ones  in  the  first 


PERMUTATIONS    AND    COMBINATIONS.  3S5 

vertical  column  are  omitted.  Hence,  2*  —  1,  espresses  the  sum  of  the 
combinations  of  t  letters,  taken  1  and  1,  2  and  2  ....  t  and  t.  Thus, 
to  apply  the  formula,  let  it  be  required  to  determine  the  number  of 
combinations  of  three  letters,  taken  1  and  1,  2  and  2,  3  and  3.  Then 
t  =3,  and  2*  —  1  =  2='  —  1  =  8  —  1  =  7.  This  agrees  with  the 
fact,  for  the  three  letters,  a,  h,  c,  give  us  the  seven  combinations,  u, 
h,  c ;  ah,  ac,  be,  and  ahr.  So,  likewise,  the  four  letters,  a,  b,  c,  J,  by 
the  formula,  give  a  number  of  combinations  equal  to  2^  —  1  =16  —  1 
=  15;  and  we,  in  fact,  have  a,  b,  c,  d ;  ab,  ac,  ad,  be,  bd,  cd  ;  ahc,  abd, 
acd,  bed ;  and  abed. 

401.  We  will  now  deduce  an  analogous  formula  to  the  above,  for  the 
sum  of  the  permutations  of  n  letters,  taken  1  and  1,  2  and  2,  3  and  3, 
n  and  ?i,  when  each  letter  is  combined  with  itself,  so  as  to  be  taken  t(j 
the  second,  third,  fourth,  and  .  .  .  n""  powers. 


PERMUTATIONS  IN  WIIICII  THE  LETTERS  ARE  REPEATED. 

Two  letters,  a  and  b,  permuted  in  this  way,  give  the  four  permuta- 
tions, aa,  ab,  ba,  bb.  Hence,  the  number  of  permutations  of  two  let- 
ters, taken  2  and  2,  when  each  letter  is  associated  with  itself,  is  equal 
to  2^.  Three  letters,  a,  b,  c,  give  the  nine  permutations,  aa,  ab,  ae,  ba, 
ea,  bb,  be,  cb,  ce.  Hence,  the  number  of  permutations  of  three  letters, 
taken  2  and  2,  is  equal  to  3^     Four  lettci-s,  a,  b,  c,  d,  give,  when  taken  ' 

2  and  2,  the  sixteen  permutations,  aa,  ab,  ac,  ad,  ba,  ca,  da,  bb,  he, 
bd,  ch,  dh,  cc,  cd,  de,  dd.  Hence,  the  number  of  permutations  of  four 
letters,  taken  2  and  2,  is  equal  to  4^.  And,  in  general,  it  is  plain  that 
the  number  of  permutations  of  n  letters,  taken  2  and  2,  is  equal  to  n^. 
We  will  next  show  that  the  number  of  permutations  of  n  letters,  taken 

3  and  3,  is  equal  to  n^.  The  three  letters,  a,  h,  and  c,  when  permuted 
in  this  way,  give  the  twenty-seven  permutations, 

aaa,  aah,  aae,  aba,  baa,  aca,  eaa,  ahc,  cba  ; 
hbh,  bbc,  bba,  bcb,  ebb,  bah,  abb,  bac,  hca  ; 
cec,  ccb,  eca,  cb.-,  bee,  eac,  aec,  cab,  aeh. 

Hence,  the  number  of  permutations  of  three  letters,  taken  3  and  3, 
is  equal  to  3^  And,  in  general,  of  n  letters,  taken  3  and  3,  the  num- 
ber is  11^.  In  like  manner,  the  number  of  permutations  of  n  letters, 
taken  4  and  4,  is  expressed  by  W*.  Hence,  for  the  entire  sum  of  the 
permutations,  taken  1  and  1,  2  and  2,  3  and  3  .  .  .  .  n  and  n,  we  have 
33  z 


38G  PERJIUTATIONS     AND     COMBINATIONS. 

,  («"  1)?!,       ,  ,      ,  .         . 

n  -\-  ir  -}-  ir -(-  n°  =  ^^ ^ — ,  the  whole,  constituting  a  geo- 
metrical series,  whose  common  ratio  is  n.  Thus,  to  apply  the  formula, 
suppose  it  be  required  to  determine  the  number  of  permutations  of  five 

n       ■,(«•'  — 1>       (5^  —  1)5       15620       „,,^,       ^ 

letters,  then  n  =  5,  and  ^ ^ —  =  ^^ -. =  — -. —  =  o'JUo,  and 

'  '  11  —  1  4  4 

/'■^^n \)7l 

if  n  =  24,  the  number  of  letters  in  the  alphabet,  we  have ~  = 

n  —  1 

(24^^^^^  =  1301724288887252999425128493402200. 

Thus,  let  it  be  required  to  determine  the  entire  number  of  changes 
that  can  be  made  with  the  three  vowels,  a,  e,  and  /.     The  formula 

^ r^— ,  becomes  -^^ — ^         =  — -  .  3  :=  39,  and  we  nnd  that  we  have 

11  —  1    '  2  2 

the  thirty-nine  permutations. 

a,  €,  t,  taken  1  and  1 ;  ae,  at,  ea,  ia,  ei,  ie,  aa,  ee,  ii,  taken  2  and  2, 
aei,  eai,  iea,  eia,  iae,  axe,  taken  3  and  3,  in  the  usual  way;  aaa,  aae, 
aai,  eaa,  iaa,  aea,  aia,  eee,  eea,  cei,  aee,  iee,  eae,  eie,  tii,  iia,  He,  aii, 
eii,  iai,  ici,  taken  3  and  3,  when  the  letters  are  combined  with  them- 
selves. 


PARTIAL  PERMUTATION. 

'  402.  Sometimes  the  nature  of  the  problem  is  such,  that  some  of  the 
quantities  cannot  be  permuted  with  each  other.  Thus,  let  it  be  re- 
quired to  determine  how  many  words  of  two  letters  each  can  be  formed 
out  of  the  letters  a,  e,  c,  d,  admitting  that  the  consonants  associated 
together  will  not  form  a  word.  The  formula  for  the  number  of  per- 
mutations of  m  letters,  taken  n  and  n,  gives  us  4  (4  —  1)  =  12.  But 
we  have  to  reject  the  number  of  permutations  of  two  letters,  c  and  (/, 
taken  2  and  2,  or  2  (2  —  1)  =  2.  Hence,  12  —  2  =  10,  the  number 
of  permutations  that  can  be  formed  in  the  required  manner.  They 
arc,  ac,  ac,  ad,  ea,  ca,  da,  ec,  ed,  ce,  de. 


RULE. 

Find  the  entire  nvmber  of  ^'""'"t^'^'totions  of  n  letters,  taken  n  and  n, 
nnd  from  this  subtract  the  number  of  permutations  of  the  p  especial 
letters,  taken  n  and  n. 


GENERAL    EXAMPLES     IN    PERMUTATIONS,     ETC.  387 


EXAMPLES. 

1.  How  many  words  of  two  letters  each  can  be  formed  out  of  the  4 
first  letters  of  the  alphabet,  assuming  that  consonants  alone  will  not 
form  a  word  ? 

A71S.  4  (4  —  1)  —  3  (3  —  1)  =  6 ;    words,  ab,  ac,  ad,  ha,  ca,  da. 

2.  How  many  numbers  composed  of  3  digits  each  can  be  formed 
from  the  number  24371,  in  such  a  way  that  none  of  the  numbers  con- 
tains only  odd  digits  ? 

Ans.  5  (5  —  1)  (5  —  2)  —  3  (3  —  1)  (3  —  2)  =  54.  Numbers, 
243,  234,  342,  324,  432,  423 ;  247,  274,  472,  427,  742,  724 ;  241, 
214,  412,  421,  142,  124;  237,  273,  372,  327,  732,  723;  231,  213, 
312,  321,  132,  123;  271,  217,  721,  712,  172,  127;  437,  473,  374, 
347,  734,  743 ;  431,  413,  341,  314,  134,  143 ;  471,  417,  741,  714, 
174,  147' 

3.  How  many  words  of  0  letters  each  can  be  formed  out  of  G  vowels 
and  6  consonants,  assuming  that  the  consonants  by  themselves  cannot 
form  a  word  ? 

Ans.  12  (12  —  1)  (12  —  2)  (12  —  3)  (12  —  4)  (12  —  5)  — 
6  (6  —  1)  (6  —  2)  (6  —  3)  (6  —  4)  (6  —  5)  =  6G45G0. 

4.  How  many  words  of  six  letters  each  can  be  formed  from  the  20 
consonants  and  G  vowels  of  the  alphabet,  assuming  that  the  consonants 
alone  cannot  form  a  word  ?  Ans.  137858400. 


GENERAL  EXAMPLES  IN  PERMUTATIONS  AND  COMBI- 
NATIONS. 

1.  Three  travellers  on  a  journey  reach  three  roads:  in  how  many 
different  positions,  with  respect  to  each  other,  may  they  continue 
travelling,  provided,  that  no  two  of  them  pursue  the  same  road  ? 

Ans.  G.  A  may  take  the  first  road,  B  the  second,  and  C  the 
third;  or,  A  the  first,  C  the  second,  B  the  third,  &c. 

2.  How  many  days  can  3  persons  be  placed  in  a  different  position 
at  dinner  ?  Ans.   G  days. 

3.  How  many  days  can  8  persons  be  placed  in  a  different  position  at 
dinner?  Ans.  40320  days. 


888  GENERAL    EXAMPLES    IN 

4.  In  how  many  different  positions  can  a  stage  driver  place  the  4 
horses  of  his  team  in  harness  ?  Aiis.  24. 

5.  How  many  changes  may  be  made  in  the  words  of  the  sentence : 
"  Lazy  boys  make  worthless,  vicious  men  ?"  Ans.  720. 

G.   How  many  changes  may  be  struck  with  10  keys  of  a  piano  ? 

Ans.  3028800. 

7.  How  many  words  of  five  letters  each  can  be  made  out  of  the  first 
24  letters  of  the  alphabet,  assuming  that  no  letter  is  repeated  more 
than  once  in  the  same  word  ?  Ans.  5100480. 

8.  How  many  words  of  three  letters  each  can  be  made  out  of  the  24 
first  letters  of  the  alphabet,  with  the  same  proviso  as  the  preceding  ? 

Ans.  12144. 

9.  How  many  numbers  of  two  digits  each  can  be  made  out  of  the 
numbers  1243  (provided,  that  none  of  the  digits  of  1243  appear  twice 
in  the  same  number),  and  what  are  they  ? 

Ans.  12.    Numbers,  12,  43,  14, 13,  21,  34,  41,  31,  24,  23,  42,  32. 

10.  How  many  numbers  of  five  digits  each  can  be  formed  out  of  the 
number  187G543,  provided,  that  no  digit  of  the  given  number  appear 
twice  in  the  required  results  ?  Ans.  2520. 

11.  How  many  changes  can  be  made  in  every  file  of  2  men  in  a 
squad  of  20  men,  so  that  the  files  shall  differ  by  at  least  one  man  ? 

Ans.  190. 

12.  How  many  changes  can  be  made  in  the  position  of  the  first  12 
letters  of  the  alphabet  ?  Ans.  479001600. 

13.  How  many  numbers  of  1,  2,  and  3  digits  can  be  formed  out  of 
the  number  476,  provided,  that  no  digit  is  repeated  in  the  isame 
number  ? 

Ans.  15.  Numbers,  4,  7,  6:  47,  46,  74,  64,  67,  76;  476,  467, 
746,  764,  647,  674 

14.  How  many  nuiubers  of  1,  2,  and  3  digits  can  be  formed  out  of 
the  number  476,  when  each  digit  is  repeated  1,  2,  and  3  times  in  the 
respective  numbers  ? 

Ans.  39.  Numbers,  4,  7,  6,  44,  47,  46,  74,  64,  67,  76,  77,  GO,  476, 
467,  746,  764,  647,  674 ;  444,  447,  446,  744,  644,  474,  464 ;  777, 
774,  776,  477,  677,  747,  767;  666,  664,  667,  466,  766,  646,  676. 


PERMUTATIONS    AND    COMBINATIONS.  380 

15.  How  many  numbers  of  1,  2,  3,  4,  and  5  digits  can  be  formed 
out  of  the  number  56789,  provided,  that  no  digit  occurs  more  than 
once  in  the  connection  with  exactly  the  same  digits  ? 

Ans.  31.  Numbers,  5,  6,  7,  8,  9 ;  56,  57,  58,  59,  67,  68,  60,  78, 
79,89;  567,  568,  569,  578,  579,  589,678,  679,  689,789;  5678, 
5679,  5689,  5789,  6789 ;  56789. 

16.  Three  travellers,  A,  B,  C,  on  their  journey  come  to 
three  roads,  and  may  either  travel  all  together  on  one  of  the 
three,  or  singly,  on  the  three  roads,  or  two  on  the  same  road, 
and  the  third  on  a  different  road.  How  many  selections  may 
they  make  of  their  routes  ? 

Ans.  27.  Three,  when  all  together;  six,  when  they  go  separately;  six, 
when  A  and  B  go  together,  and  C  goes  by  himself;  six,  when  A  and 
C  are  together,  and  B  by  himself;  six,  when  B  and  C  are  together, 
and  A  by  himself. 

17.  IIow  many  numbers  composed  of  three  digits  each  can  be  made 
out  of  the  number  123456780,  provided,  that  all  the  numbers  thus 
formed  shall  differ  by  at  least  one  digit.  Ans.  84. 

18.  How  many  numbers  of  4  digits  each  can  be  formed  out  of  the 
same  number,  128456780,  so  as  to  fulfil  the  above  condition  ? 

Ans.  126. 

10.  How  many  numbers  composed  of  single  digits,  of  2  digits,  of  3 
digits,  of  4  digits,  and  of  5  digits,  can  be  formed  out  of  the  number 
12345,  in  such  a  way  that  each  number  shall  differ  from  all  the  other 
numbers  by  at  least  one  digit. 

A71S.  31.  Numbers,  1,  2,  3,  4,  5 ;  12,  13,  14,  15,  23,  24,  25,  34, 
35,  45;  123,  124,  125,  134,  135,  145,  234,  235,  245,  345;  1234, 
1235,  1345,  1452,  2.345 ;  12345 

20.  How  many  numbers  composed  of  1,  2,  3,  4,  and  5  digits  can  be 
formed  out  of  the  number  12345,  in  such  a  way  that  the  same  digit 
may  be  repeated  once,  twice,  &c.,  in  the  same  number,  and  the  several 
numbers  necessarily  differing  only  in  the  position  of  the  digits. 

Ans.  3005.  Numbers,  1,  2,  3,  4,  5;  12,  13,  14,  15,  21,  31,  41, 
51,  23,  24,  25,  32,  42,  52,  34,  35,  43,  53,  45,  54,  11,  22,  33,  44, 
55;  111,112,  113,  114,  115,  211,  311,411,  511,  121,  131,  141, 
151,  &c. 

33* 


390  LOGARITHMS. 

21.  A  gentleman  being  asked  what  he  would  take  for  a  valuable 
horse,  replied,  that  his  price  was  a  cent  for  every  change  that  he  could 
make  in  the  position  of  the  32  nails  in  the  horse's  shoes.  What  was 
his  price? 

A71S.  2'631308S69336'93530l'672180121600600  dollars. 


LOGARITHMS. 

403.  Tlie  logarithm  of  a  quantity  is  the  exponent  of  the  power  to 
which  it  is  necessary  to  raise  an  invariable  quantity,  called  the  base,  to 
produce  the  quantity. 

Let  a  be  the  invariable  quantity  or  base,  x  its  logarithm,  and  i/  the 
quantity  given ;  then,  a^  =  y.  In  this  equation,  x  is  the  logarithm 
of  1/.  It  is  usual  to  write  logarithm,  log.,  or  simply,  1.  Hence, 
X  =  log.  y,  or  1.  1/.  We  will  begin  the  discussion  by  supposing  a  ^  1, 
and  X  positive,  and,  though  the  general  principles  of  logarithms  are  true 
for  quantities  as  well  as  for  numbers,  yet,  as  iu  practice  the  logarithms 
of  algebraic  quantities  are  seldom  used,  we  will  first  show  the  relations 
between  numbers  and  their  logarithms.  Resuming  the  equation,  a*  =  y, 
we  see  that  when  i/  =  1,  x  =  0,  for,  a"  =:  1.  Hence,  log.  1  =  0,  and, 
since  this  will  be  so,  whatever  may  be  the  value  of  a,  we  see  that  the 
log.  1  is  always  zero.  Every  value  given  to  x,  above  zero,  will  cause 
1/  to  increase  more  and  more ;  when  a;  =  1, ,?/  =  a.  When  x  has  values 
attributed  to  it  greater  than  1,  y  will  exceed  a  more  and  more,  until 
x  =  CO  gives  also  y  =  oo  .  Suppose,  for  instance,  a  =  8,  then 
x  =  0  and  x  =  1  give  y  ==  1  and  y  =  S.  Moreover,  x  =  1,  2,  3, 
4,  &c.,  gives  1/  =  8,  64,  512,  4096,  &c.  Finally,  a;  =  oo  gives  y  =  go  , 
for,  8^"  =  CO  .  It  is  plain  that,  when  x  has  an  intermediate  value 
between  0  and  1,  that  y  will  have  a  value  between  1  and  8, 
and  that  by  making  x  pass  through  all  possible  positive  values, 
entire  and  fractional,  between  0  and  1,  and  1  and  oo  ,  y  will  be  made 
to  pass  through  all  possible  positive  values  between  1  and  8,  and 
8  and  cc  .     The  contrary  of  what  we  have  just  seen  will  be  the  case 

when  X  is  negative.  For  the  equation,  a~''  =y,  will  become  —  =  y;  and 

it  is  plain  that,  as  x  increases  y  will  decrease,  and  that  when  x  =  cc  , 
y  =  0.     To  illustrate  this,  let  a  =  8,  and  let  cr  r=:  0,  1,  2,  3,  4,  &c., 


LOGARITHMS.  391 

then,  y  =  1,  |,  g'j,  ^jj,  4 g^g,  &c.  And  it  is  evident  that,  by  giving 
to  X  all  possible  positive  values,  entire  and  fractional,  between  U  and  oo  , 
y  may  be  made  to  pass  through  all  possible  positive  values  between  1 
and  0.  Suppose  now,  a  <^  1.  It  will  still  be  true  that  all  possible 
values  may  be  formed  from  the  powers  of  one  number,  but  the  order  of 
the  numbers  will  be  reversed.  For  values  of  ar,  between  the  limits  0 
and  —  CO ,  we  will  get  all  possible  positive  numbers  between  1 
and  +  00  ;  and  for  all  values  of  x,  between  0  and  +  oc  ,  we  will  get 
all  possible  positive  numbers  between  1  and  0. 

The  remarkable  fact  that  all  positive  numbers  might  be  regarded  as 
the  powers  of  one  invariable  number,  led  to  the  invention  of  logarithmic 
tables  by  Napier,  for  the  purpose  of  abridging  complicated  numerical 
calculations.  The  invariable  number  is  called  tlie  base  of  the  si/sfem  of 
logarithms,  and  may  evidently  be  any  number  whatever,  except  unity. 
Since  all  the  powers  of  unity  are  unity,  it  is  plain  that  this  number 
cannot  be  taken  as  a  base.  The  base  of  the  common  system  of  loga- 
rithms is  10.  The  powers  of  this  number  are  more  readily  formed 
than  the  powers  of  any  other  number  whatever,  and  this  was  the  main 
reason  for  its  selection. 

We  have  (10/  =  1,  Hence,  log.  1        =0, 

(10)' =  10,  ''       log.  10      =.1, 

(10)^=100,  "       log.  100    =2, 

(lOy  =  1000,  "      log.  1000  =  3, 

&c.  &c. 

We  see  that,  as  the  number  changes  (the  base  being  the  same),  the 
logarithm  also  changes.  It  is  plain,  moreover,  that  any  change  in  the 
base,  the  number  being  unaltered,  will  produce  a  change  in  the  loga- 
rithm. Thus,  if  the  base  is  12,  since  12'  =  12,  we  have  log.  12  :=  1. 
The  logarithm  of  10  in  this  system  will  then  not  be  1,  as  in  the  system 
whose  base  is  10,  but  will  be  some  fractional  number  less  than  1. 
Hence,  we  see  that  logarithms  depend  upon  the  number  and  upon  the  i 
base.  That  part  of  the  logarithm  in  any  system  which  depends  upon 
the  base  is  called  the  modulus  of  the  system.  A  table  of  loijarithms  is  a 
table  exhibiting  the  logarithms  of  all  numbers  between  certain  limits, 
as  0  and  100,  or  0  and  10000,  calculated  to  a  particular  base. 

The  general  properties  of  logarithms  are  independent  of  any  particular  j 
base.     A  few  of  these  general  properties  will  now  be  demonstrated. 


392  LOGARITHMS. 


First  Property. 

404.  The  logarithm  of  the  product  of  any  number  of  factors  taken 
in  the  same  system,  is  equal  to  the  sum  of  the  logarithms  of  those 
factors. 

For,  let  a  represent  the  base  of  the  system,  y,  y',  ?/',  y'",  &c.,  several 
numbers,  and  x,  x',  x",  x'",  &c.,  their  corresponding  logarithms.  Then, 
cv^  =  y,  o"'  =:  y' ,  a""  =  y",  a"""  =  y"',  &c.  Multipl3'ing  these  equa- 
tions together,  member  by  member,  we  have  0^+^'+""+^'"+ ^'-  z=i  y  ij y" y'" , 
&c.  Then,  since  the  exponent  of  the  base  is  (from  the  definition)  the 
logarithm  of  the  second  member,  we  have  x  +  x'  +  x"  +  a;"'  +  &c. 
=  log.  yy'y"^",  &c.  But  the  first  equations  give  x  =  log.  y,  x'  = 
log.  y',  x'  =  log.  y",  &c.  Hence,  log.  y  -\-  log.  y'  +  log.  y"  +  log  :!/"' 
+  &c.  =  log  yy'y"y"'  &c.,  as  enunciated.  Then,  since  the  logarithm 
of  the  product  of  any  number  of  factors  is  equal  to  the  sum  of  the 
logarithms  of  these  factors,  it  follows  that  to  multiply  by  means  of  loga- 
rithms, we  have  only  to  add  together  the  logarithms  of  the  factors,  and 
to  find  the  number  corresponding  to  this  sum.  This  number  will  be 
the  product  required.  Thus,  let  it  be  required  to  multiply  100  by  10, 
by  means  of  logarithms.  The  log  of  100  is  2,  in  a  system  whose  base 
is  10 ;  the  log  of  10  is  1 ;  the  sum  of  these  logarithms,  3,  is  the  loga- 
rithm of  the  product.  Look  for  the  number  corresponding  to  3  as  a 
logarithm,  and  you  will  have  1000  as  the  product  required.  To  exhibit 
the  whole  in  an  equation,  we  have,  log.  100  x  10  =  log.  100  -f  log.  10 
=  2-1-1  =  3;  the  number  corresponding  to  which  is  1000. 

We  have,  then,  for  multiplication  by  means  of  logarithms,  this 

RULE. 

Add  toycther  the  logarithms  of  the  several  factors.  This  sum  will  he 
equal  to  the  logarithm  of  the  product.  Looh  in  the  table  of  logarithms 
for  the  number  corresponding  to  this  logarithm,  and  you  loill  have  the 
product  required. 


1.  Required  the  equivalent  expression  to  log.  a   b   c   d. 

Ans.  Log.  a  +  log.  b  -\-  log.  c  -f-  log.  d. 

2.  Required  the  equivalent  expression  to  log.  (San. 

Ans.   Log.  6  -f-  log.  a  -f-  log.  n. 


LOGARITHMS.  393 

3.  Required  the  equivalent  expression  to  log.  (a  +  by  c"  d^. 

Ans.  Log.  (a  +  i)"  +  ^og.  c"  +  log.  d^. 


Second  Projperty. 

405.  The  logarithm  of  the  quotient  arising  from  dividing  one  quan- 
tity by  another  is  equal  to  the  logarithm  of  the  dividend  minus  the 
logarithm  of  the  divisor.      For,  let  x  equal  the  quotient  of  two  quan- 

11  ^''^  4  1-1 

titles,  m  and  n,  then  x  =  — ,  or   nx  =  m.       And,   since,   when    two 
n 

quantities  are  equal  their  logarithms  must  be  equal,  we  have  log.  nx 
=  log.  m,  or  log.  n  +  log.  x  =  log.  w.  Hence,  log.  x  =  log.  m  —  log. 
n,  as  enunciated.  From  this  we  derive  for  division,  by  means  of  loga- 
rithms, the  following 

RULE. 

From  the  logarithm  of  (he  dividend  subtract  the  logarithm  of  the 
divisor;  the  remainder  will  be  the  logarithm  of  the  quotient.  The 
number  in  the  table  corresponding  to  this  logarithm  will  be  the  quotient 
required. 

Thus,  to  find  the  quotient  of  1000  by  10,  we  have  log.  1000  —  log.  10 
=  3  —  1  =  2,  and  the  number  corresponding  to  2  as  a  logarithm  is 
100. 


1.  Find  the  equivalent  expression  to  log.  - — . 

bn 

Ans.  Log.  am  —  log.  bn,  or  log.  a  +  log.  m  —  log.  b  —  log.  n. 

2.  Find  the  equivalent  expression  to  log.  ^ — ~. 

Ans.  Log.  X  -f-  log.  ij  —  log.  (b  -\-  n). 


Find  the  equivalent  expression  to  log  "^ 


bn 
Ans.  Log.  (x  -f  _y)  —  log.  b  —  log.  n. 


4.  Find  the  equivalent  expression  to  log.  j^ 


bn 
Ans.  Log.  X  -f  log.  y  —  log.  b  —  log.  n. 


394  LOGARITHMS. 

5.  Find  the  equivalent  expression  to  log.  (y^  )• 

Ans.  Log.  (.'r?/)P  —  log.  (in)p. 

G.  Find  the  equivalent  expression  to  log.  (- V 

Ans.  Log.    («  +  i)  —  log.  c. 


Third  Property, 

406.  The  logarithm  of  the  power  of  a  number  is  equal  to  the  expo- 
nent of  the  power  into  the  logarithm  of  the  number. 

For,  take  the  equation,  as=y,  and  raise  both  members  to  the  m*" 
power,  we  will  have  a"^  =  3/°  (A).  In  equation  (A),  m  is  any  number 
whatever,  positive  or  negative,  entire  or  fractional.  But,  since  the 
exponent  of  the  base  is  the  logarithm  of  the  second  member,  equation 
(A)  gives  mx  =  log.  ?/™ ;  and  the  first  equation  gives  x  =  log.  y. 
Hence,  m  log.  y  =  log.  ?/■",  or  log  y"^  =  m  log.  y,  as  enunciated.  It 
follows,  then,  that  to  raise  a  number  to  a  power  by  means  of  logarithms, 
we  have  only  to  apply  this 

RULE. 

Multiply  the  logaritlim  of  the  numher,  as  found  in  the  tahic,  hy  the 
exponent  of  the  poioer  to  which  it  is  to  be  raised.  This  product  xciJl  he 
the  logarithm  of  the  poioer,  a7id  the  number  found  in  the  table  corres- 
ponding to  this  logaritlim  loill  he  the  required  poioer. 

Thus,  to  raise  10  to  the  second  power  by  means  of  logarithms,  we 
multiply  the  logarithm  of  10,  which  is  1,  by  2,  the  exponent  of  the 
power.  The  product,  2,  is  the  logarithm  of  the  power,  and  the  number 
corresponding  to  this  logarithm  is  100.  So  that  we  have  log.  (10)^  = 
2  log.  10  =  2 ;  the  number  corresponding  to  which  is  100. 


1.  Find  the  equivalent  expression  to  log.  (a  +  bj^. 

Ans.  m  log.  (a  H-  b). 

2.  Find  the  equivalent  expression  to  log.  c"  (a  +  t)'". 

Ans.  n  log.  c  +  m  log.  (a  +  b). 
The  1st.  property  is  here  used  in  connection  with  the  3d. 


LOGARITHMS. 


3.  Find  the  equivalent  expression  to  log.  a"  {mry. 

Alls,  n  (log.  a  +  log.  r  +  log.  m). 

4.  Find  the  equivalent  expression  to  log.  a»  (-A  . 

Ans.  11  (log.  a  +  log.  h  —  log.  c). 
The  2d.  and  3d.  properties  both  employed. 

5.  Find  the  equivalent  expression  to  log.  ^^ — j^ —  . 

Ans.  m  log.  (a  -\-  I)  —  2  log.  h. 

6.  Find  the  equivalent  expression  to  log.  — ^-g. 

Ans.  Log.  (a  —  x)  —  2  log.  (a  -j-  x). 

(a  -f  xY 

7.  Find  the  equivalent  expression  to  log.  — ^ ^. 

Ans.  2  log.  (a  +  x)  —  log.  (a  —  x). 


Fourth  Properti/. 

407.  The  logarithm  of  the  root  of  a  number  is  equal  to  the  logarithm 
of  the  number  divided  by  the  index  of  the  root. 

For,  take  the  equation,  a^  =  y,  and  extract  the  m""  root  of  both 

X  3j  

members.     Then,  Va"  =  '</,?/,  or,  a^   ='y/y-     Hence,  —  =log.  Vy 

(A.),   the   exponent  of  the  base  being  the  logarithm  of  the  second 
member.     But  the  first  equation  gives  x  =  log.  y  ;  equation  (A)  then 

becomes  log.  ^—  =  log.  "V'y,  or  log.  ly  v  =  log.  ^—,  as  enunciated. 

in  "-    m 

For  extracting  roots  by  means  of  logarithms  we  have,  therefore,  this 


RULE. 

D'lcide  the  logarithm  of  the  number,  lohose  root  is  to  be  extracted,  by 
the  index  of  the  root.  The  quotient  toill  be  the  logarithm  of  the  root; 
the  number  found  in  the  table,  corresponding  to  this  logarithm,  will  be 
the  root  required. 

Thus,  to  extract  the  square  root  of  100  by  means  of  logarithms,  we 
divide  2,  the  logarithm  of  100,  by  2,  the  index  of  the  root.     The  quo- 


396 


L  0  G  A  R I  T  H  JI  S 


tient,  1,  is  the  logarithm  of  the  root,  aud  10,  the  number  con-espoudiug 
to  this  logarithm,  is  the  root  required. 


EXAMPLES. 

1.  Find  the  equivalent  expression  to  log.  ^~x  +  xj. 

.  Log.  (.-:•  +  y^ 


2.  Find  the  equivalent  expression  to  log.  ^d^  —  .r^. 

Log,  (g^-x^)        log,  (g  +  .T)  („  -  ,:) 

^LUS.  ^  —   _ :^    1    log.    (^a    ^   x) 

+  i  log.  (a  —  x). 

3.  Find  the  equivalent  expression  to  log.  a^  ^a^. 

Ans.  -g-  log.  a. 
o 

4.  Find  the  equivalent  expression  to  log.  ^/(o  +  by. 

Ans.  I  log.  (a  +  F). 

408.  Since,  a°  =  1,  it  matters  not  what  is  the  value  of  a,  and,  since  the 
equation,  a°  =  1,  gives  0  =  log.  1,  it  follows  that  the  logarithm  of  unity 
in  every  system  of  logarithms  is  equal  to  zero.  From  this  we  readily 
deduce  the 

Fifth  Property. 

The  logarithm  of  the  reciprocal  of  a  number  is  equal  to  the  loga- 
rithm of  the  number  taken  negatively. 

For,  let  N  bo  the  number,  then  -r^  will  be  the  reciprocal  of  the  num- 
ber, and  we  have  log.  ^  =  log.  1  —  log.  N  =  0  —  log.  N  =  —  log.  N, 
as  enunciated. 

EXAMPLES. 

(a  4-  xY 
1.  Required  the  equivalent   expressions  to  log.   1^ ^—-,  and  log. 


(a  -f  xf 

Ans.  2  log.  (a  -f-  a")  —  3  log.  x,  and  3  log.  x  —  2  log.  (a  -{■  a;). 


LOGARITHMS.  397 

(a- r^) 

2.  Eequired    equivalent   expressions   to   log.    ^ ~,    and    log. 

(.a  +  xy 

a'  —  x'' 
Ans.  Log.  (a  —  x)  —  3  log.  (a  + j:),and31og.  (cy+.t)  —  log.  («  —  x). 


Corollary. 

409.  "When  a  ^  1,  we  have  a"  =  oo  .  Hence,  oo  =  log.  x  .  That 
is,  the  logarithm  of  infinity  in  a  system,  whose  base  is  greater  than  unity, 

is,  itself,  infinity.    And,  since  0  =  — ,  it  follows  from  the  fifth  property, 

that  log.  0  =  —  -x.  .  That  is,  the  logarithm  of  zero  in  a  system,  whose 
base  is  greater  than  unity,  is  minus  infinity. 

When  a  <^  J^,  a  number,  y^  for  example,  it  is  plain  that  the  base 
must  be  raised  to  a  power  denoted  by  —  oc,  in  order  to  produce  in- 
finity.    Thus,  (j'jj)""  =  r-f^  =  (10)=^  =  X  .     Hence,  the  logarithm 

of  infinity  in  a  system,  whose  base  is  less  than  unity,  is  minus  infinity. 
Hence,  by  the  fifth  property,  the  logarithm  of  zero  in  a  system  whose 
base  is  less  than  1,  is  plus  infinity.  Thus,  it  is  plain  that  (jjj)°°  =  0, 
from  which   oo  =  I02;.  0. 


Sixth  Propertij. 

■110.  The  logarithm  of  any  base,  taken  in  its  own  S3-stcm,  is  unity. 

For,  suppose  there  are  any  number  of  bases,  a,  h,  c,  &c.,  and  desig- 
nate the  logarithms  in  the  systems,  of  which  these  are  the  bases,  by 
log.  log.'  log."  &c.  We  have  a'  =  a,  i'  =  h,  c'  =  c,  &c.  Hence, 
1  =  log.  a,  1  =  log.'  6,  1  =  log.'  c,  &c. 

Thus,  1  is  the  logarithm  of  10  in  a  system  whose  base  is  10.  But, 
in  the  system  whose  base  is  8,  the  logarithm  of  10  must  be  greater 
than  1.  Because,  8'  =  8,  or  1  =  log.  8.  Hence,  a  number  greater 
than  1  must  be  the  logarithm  of  10. 

Seventh  Principle. 

411.  If  we  have  a  table  of  logarithms  calculated  to  a  particular  base, 
the  logarithms  of  the  numbers  in  this  table,  divided  by  the  logarithm 
of  a  second  base,  taken  in  the  first  system,  will  give,  as  quotients,  the 
logarithms  of  the  same  numbers  in  the  second  system. 
34 


SnS  T.OGAIUTIIMS. 

For,  let  fl*  =;/,  and  i^  =  ?/.     Then  a*  =  i'-  (A).     Distinguish  the 

logarithms  of  the  two  systems  by  log.  and  log.'.     Taking  the  log.  of  both 

members  of  (A),  we  have  x  log.  a  =  z  log.  h.     But,  since  x  =  log.  y, 

and  z  =  log'.  ;y,  and  log.  a  =1,  this  equation  becomes  log.  1/  =  log.'y, 

lo"".  1/ 
lotr.  i.     Then  log.'  y  =  ,  ^  '  ,  as  enunciated. 
^  ^    "^        log.  6 

412.  The  two  systems  in  most  common  use  are  the  Napierian,  whose 
base  is  2.718281828,  and  the  common,  whose  base  is  10.  Suppose 
the  Napierian  logarithms  to  be  designated  by  log.',  and  the  common  by 
log.  Then,  if  we  knew  the  common  logarithm  of  any  number,  we  can 
get  the  Napierian  logarithm  of  the  same  number,  by  dividing  the 
common  logarithm  of  the  number  by  the  common  logarithm  of  the 
Napierian  base;  or  we  can  get  the  common  logarithm  of  a  number, 
knowing  its  Napierian  logarithm,  by  multiplying  the  Napierian  logarithm 
by  the  common  logarithm  of  the  Napierian  base. 


DIFFERENTIAL  OF  a'. 

413.  The  Binomial  Formula  enables  us  to  find  the  differential  of  a'. 
For,  let  u  =  a^,  and  give  x  an  increment,  h.  Then,  tt'  =  a'^^^  =  a'^a}', 
and  u'  —  u  =  a^^a}"  —  1).      Let  a  =  1  -}-  b,  and  develop  o''  by  the 

binomial  fonnula,  we  will  have  a^  =  (1  -f  &)''  =  1  +  hh  +  -^ — -^^ 

+  'ii^2M*^^.  ..  ^  ,.  +  .(„_|  +  |_|  +1). 

by  neglecting  the  higher  powers  of  h,  as  infinitely  small  quantities  of 
the  second  order  and  higher  orders.     But,  since  h  =  a  —  1,  we  have 

a"  =  1  +  /i(  (a  —  1)  — ^^iizl^'  +  illzlT  —  Sic.)  =  1  +  mi,  by 

designating  the  quantity  within  the  parenthesis  by  N.  Hence,  u'  —  u 
z=a^(ci^  — 1)  =  a'^N/i,    and    du  =  a^'Ndx  =  iiNdx;    from    which 

dx  =  -^.  — ■     And,  since  u  =  a^,  we  have  x  =  log.  m,  and  dx  = 

N       M  '' 

(^(log.  «).  Moreover,  since  a  is  the  base,  and  N  depends  entirely  upon 
a,  we  see  that  the  differential  of  a  logarithm  is  equal  to  the  reciprocal 
of  a  constant,  dependent  upon  the  base  into  the  differential  of  the 
quantity  divided  by  the  quantity. 


LOGARITHMS.  61 

EXAMPLES. 

1  Required  differential  of  b^.  Ans.    Ni^Y/i/. 


2.  Required  differential  of  log.  (1  +  re). 

1    ^(l+x)_l 

N'     1  +  X    ~N'l  +  x' 

3,  Required  the  differential  of  log.  (1  +  xy. 

Ans. 


dx 


dx 


N      (\  +  xf  N  ■  (1  +  x)' 


LOGARITHMIC  SERIES. 

414.  A  logarithmic  series  is  one  which  will  enable  us  to  calculate 
approximatively  the  logarithms  of  any  number  whatever.  We  have  seen 
that  the  logarithm  of  a  number  depends,  1st,  on  the  number ;  2d,  on 
the  base.     Hence,  the  development  of  a  logarithm  must  contain  the  \ 
number,  or  some  quantity  dependent   upon    the    number,  and   some  / 
quantity  dependent  upon  the  base.     These  two  facts  serve  as  a  guide 
in  assuming  the  form  of  development.     Suppose  we  have  the  equation, 
a'  =  x,  in  which  x  is   the   number,  and  i/  the   logarithm.     Let   us 
assume  log.  a;  =  A  +  A'.r  -|-  A'V  +  A"'x'  +  &c.,  in  which  A,  A', 
A",  &c.,  arc  independent  of  x,  and  dependent  upon  the  base,  a.    Now, 
if  we  make  .r  =  0,  the  first  member  becomes  infinite,  whilst  the  second 
reduces  to  a  finite  quantity,  A.     The  assumed  form  is  then  wrong. 
Again,  assume  log.  x  =  Kx  +  A'x  +  A'V  +  &c.      Making  x  =  0, 
we  get  ±  Qo  =  0,  which  is  absurd.     The  development  then  cannot  be 
made  under  the  second  form.     An  examination  of  the  two  forms  shows 
that  we  have  assumed  the  logarithm  to  be  developed  in  the  powers  of 
the  number,  and  we  have  seen  that  it  cannot  be  so  developed.     Since 
log.  x  cannot  be  directly  developed,  let  us  see  whether  log.  (1  +  x)  can    ^V, 
be   expanded   into  a  series.^    Assume  log.  (1  +  a;)  =  Ax  +  AV  +       f 
A'V  +  A'"*^  +  &c.  (B).     When  a?  =  0  we  have  log.  1  =  0,  wliich       ' 
presents  no  absurdity.    Taking  the  differentials  of  both  members  of  (B), 

and  dividing  out  by  dx,  we  have,  —  .  .j =  A  +  2A'.e  +  3A'V  + 

4A"V  +  &c.  (C),  in  which,  N  =  («.  —  l^  —  ^^^-ZH  4-  ^"  ~  ^^' 
—  ^c.     Making  a;  =  0  in  (C),  we  get  A  =  ^.     Differentiating  (C), 


400  LorxAHITIIMS. 

we  get,  after  dividing  out  by  ilr,  —  ^  .  \.^=  2A'  +  2  .  oA"x-  + 

JN     (i  -f-  -t) 

8  .  4  A"V  +  Sec.  (D),  whence  A'  =  —  ^^.     DiflFercutiating  (D),  and 

2  1 

dividing  out  by  dx,  we  have  +  -^  .  -r^  =  2  .  3 A" +  2  .  3  .  4A"'x 

-{■  &c.  (E).  Whence,  by  making  x  =  0,  there  results,  xV  =  -j-  ttju^- 
The  hiw  for  finding  the  coefficients  is  now  plain  ;  A'"  will  be  found  equal 

to  —  -r<^,  A'-'*'  =  -^^r^,  A^  =  —  T7T7,  &c.  lleplacing  the  constants 

4N  5N '  6N'  i-        & 

1    /  X^  X? 

in  (B)  by  their  values,,  we  have  log.  (1  +  a:)  =  -^^  (  x  —  —  +  -^  — 

T  +  l-H  <^-) 

Since  (1  +  ic),  the  number  whose  logarithm  is  developed,  has  been 
expanded  in  the  powers  of  x  ;  we  conclude  that,  though  the  logarithm 
of  a  number  cannot  be  developed  in  the  powers  of  the  number,  it  may 
be  in  the  powers  of  a  number  less  by  unity. 

Since  N  =  (a  —  1)  — ^^ — - — -  +  - — ^ — -,  &c.,  in  which  a  is  the 

base  of  the  system,  it  is  plain  that  the  factor,  — ,  in  equation  (P),  depends 

for  its  value  upon  the  base  alone.  Equation  (P)  shows,  moreover, 
that  the  logarithm  of  a  number  is  composed  of  two  factors,  one 
dependent  upon  the  base  alone,  and  the  other  dependent  upon  the 
number  alone.  The  factor  which  depends  upon  the  base  for  its  value 
is  called  the  modulus  of  the  system .  The  modulus  is  usually  represented 
by  M,  or  M'. 

415.  If  we  take  the  logarithm  of  (1  +  ic),  in  a  system  whose  base 
is  a' ,  and  denote  the  new  logarithm  by  log.',  it  is  plain  that  the  new 
development  will  differ  from  the  first  only  in  this  notation  and  in  the 
}nodulus.     Hence,  we  will  have 

log.  (1  +  ..:)  --=  M(r  _  l'  +  f'  -  |V  -f  -  &c.),  and 

log.'  (1  +  .0  =  M'(.-  -  I  +  ^-^  +  ^-  &c.). 
Hence,  log.  (1  +  x)  :  log.'  (1  +  x)  :  :  M  :  M'. 
That  is,  the  logarithms  of  the  same  mtmher  in  tioo  different  systems 
are  to  each  other  as  the  modtdi  of  these  systejns.  • 


LOGARITHMS.  401 

416.  Since  the  modulus  is  expressed  in  terms  of  the  base,  it  follows 
that  if  the  base  be  given,  the  modulus  can  be  determined,  and  con- 
versely, the  base  can  be  found  from  the  modulus. 

If,  then,  we  make  M'  =  1,  the  base  of  the  system  will  become  knowu 
The  system  of  logarithms,  whose  modulus  is  unity,  is  called  Napierian 
from  Lord  Napier,  the  inventor  of  logarithms.  The  base  of  the  system 
calculated  from  the  assumed  modulus,  is  2-718281828. 


COMMON   AND  OTHER  LOGARITHMS  FOUND  FROxM 
NAPIERIAN. 

417.  The  proportion,  log.  (I  +  x)  :  log'.  (1  +  a;)  :  :  M  :  M',  be- 
comes, when  M'  =  l,  log.  (1  +  x)  :  log'.  (1  -f  a:-)  :  M  :  1.  Hence,  we 
have  log.  (1  +  »•)  =  M  log'.  (1  -f  x)  (A). 

This  equation  shows  that  the  Napierian  logarithm  of  any  number, 
muUiplied  hy  the  modulus  of  any  other  system,  loill  give  the  logarithm 
of  the  same  numher  in  that  system.  Moreover,  equation  (A),  compared 
with  the  7th  Property,  shows  that  the  viodulus  of  any  system  is  equal  to 
the  logarithm  of  the  Napierian  base  of  that  system. 


MEASURE   OF  ANY  MODULUS. 

418.  Let  a  be  the  base  of  any  system,  and  let  the  logarithms  of  this 
system  be  designated  by  log.,  and  those  of  the  Napierian  by  log'.  Let 
(1  +  x)  be  any  number,  then  the  7th  Property  will  give  log.  (1  -i-  x) 

=  - — J —  log'.  (1  +  x)  (B).     Equation  (B)  compared  with  (A)  shows 

that  M  =  , — } — .     Hence,  the  measure  of  the  modulus  of  any  system 

is  equal  to  the  reciprocal  of  the  Napierian  logarithm  of  the  base  of  that 
system. 

This  measure  is  even  true  for  the  Napiarian  modulus  itself,  for  we 

1  1        , 

have  M  =  , — -. —  =  -  r=  1. 
log',  e         1 


TABLE   OF  NAPIERIAN  LOGARITHMS. 

419.  The  formula,  ?  (1 -f  a:)  =  M  (.T  —  "^  +  ^ — '^  +  ^ — ^ -f- &c.,) 
84*  2a 


t02  LOGARITHMS. 


X  X  X^ 

becomes,  wlien  M  =  1,  I'{1  +  x)  =  (x — -^- +  .,  — j 


+  &c.,)(C). 

Making  x  =  1,  we  have,  ^2  =  1  —  i  +  J  —  i  +  i  —  i  +  &c. 

This  series  enables  us  to  calculate  the  Napierian  logarithm  of  2,  and, 
by  making  x  =  2,S,  4,  5,  &c.,  we  can  calculate  the  Napierian  logarithm 
of  3,  4,  5,  6,  &c.  But  the  series  does  not  converge  rapidly  enough, 
because,  for  every  number  greater  than  2,  the  series  goes  on  increasing 
continually,  and  it  would  be  necessary  to  take  an  infinite  number  of 
terms  in  order  to  make  the  calculation  in  any  degree  correct.  A  simple 
artifice  will  enable  us  to  convert  the  above  series  into  a  converging  one, 
that  is,  into  a  series  in  which  the  terms  will  become  smaller  and  smaller; 
so  that  all  after  a  certain  number  may  be  neglected  without  aifecting 
the  result  materially.  Two  transformations  are  used  to  aficct  this  con- 
vergence. 

First   Transformation. 
420.   Make  x  =z  -  \\\  equation  (C),  it  will  become 

-^+&c.,(D). 

This  series  will  become  more  converging  as  y  increases. 
Making  y  =  1,  2,  3,  4,  5,  6,  &c.,  in  succession,  we  get 

>2  — n=r2  =  i  — ^  +  J  — J  +  i  — i  +  &c., 

Z'4  =  Z'3  +  i- J3  +  hV- 3k  +  T2V7r-&c., 
/'5  =  Z'4  +  1  —  3^.  +  ih  —  TTR4  +  i,~ho  —  &c- 

The  first  series  enables  us  to  calculate  the  Napierian  logarithm  of  2. 
The  second  series  enables  us  to  calculate  the  Napierian  logarithm  of  3 
when  that  of  2  is  known ;  and  the  formula,  evidently,  only  can  be  used 
when  the  logarithm  of  the  next  lowest  number  is  known. 


Second    Transformation. 

421.  The  preceding  formula  does  not  give  a  sufficiently  converging 
series  for  small  numbers.     A  better  series  can  bo  obtained  in  the  fol- 


LOGARITHMS.  403 

lowing  manner.      In  the  equation,  ?'(!  +  x)  ^=x  —  tt  -^   o r  + 

x^        x^ 

— ^4-  &c.,  we  get,  by  changing  +  .r  into  —  x,  /'(I  —  a-)  =  —  x 

3,2         ^3         ^4  j.a         ^.6 

—  — -^ -. ^  —  -; &c.     Subtracting   the   second   series 

2         3         4         5         0  ° 

from  the  first,  we  get  l'{l  -f  x)  —  ?'(1  —  .r)  =  I'L  "*"  ^\  =  2  (x  + 

^  +  ?  +  T  +  '^  +  n  +  S  +  *°-')(*=)-  p'-l^  =  i  + 

-,  in  which  z  is  a  whole  number.     Then,  x  =  ~ =-,  and  replacing  x 

in  (E)  by  this  value,  we  get  ^'(1  +  -),  or  I' (I  +  z)  —  l' z=2  (- - 

verges  more  rapidly  than  (D),  and  will  answer  even  for  small  numbers. 
Let  z  =  l,2,  o,  4,  5,  6,  &c.,  we  will  have 

re  -  «  =  2  (i  +  -1^  +  ^^  +  ^-2_  +  _2_  +  fe. 

f6  — ?5  =  2(T'r  +  8(ir)+5Tn)»+f(Tl)'+9(TT7+ll(nv' 
+  &c.) 

1.  For  small  numbers  it  is  necessary  to  take  a  good  many  terms  of 
this  series,  but  when  the  numbers    are  large  one  or  two  terms  will 

answer.  Thus,  let  a- =  1000,  then  nOOl—nOOO =2(5  J^-j-  +  o-oiTr-.^., 
+  FrF^jK(yf\5  +  ^cV     We  see   that  the   fi;-st  term  of  this  series  will 

only  add  about  jq^^o  *°    *^^  logarithm  of  1000,  and  the  second  term 
will  add  less  than   godoFoo^oo-      And,  since   logarithms  are  seldom 


404  LOGARITHMS. 

carried  further  than  the  Tth  place  of  decimals,  all  the  terms  after  the 
first  may  be  neglected. 

2.  In  constructing  a  table  it  is  only  necessary  to  calculate  the  loga- 
rithms of  prime  numbers,  since  the  logarithm  of  any  number  made  up 
of  factors  is  equal  to  the  sum  of  the  logarithms  of  its  factors.  Thus, 
no  =  n  +  I'b,  Z'12  =  Z'3  +  Z'4  =  Z'3  +  2Z'2,  &c. 

3.  Knowing  the  Napierian  logarithm  of  10,  we  can  fiud  the  modulus 
of  the   common   system.       For,  log.'  10  :  log.  10  :  :  1  :  M.       Hence, 

M  =  -^rrcT  '  ^^^  ^*-*S-  1^  "=  ^^  ^"^^  *'^®  calculated  value  of  log.'  10  is 

2.302585093.      Therefore,  M  =  ^  oaolo^ac,^  =  0-434294482. 

If  we  now  multiply  the  Napierian  logarithms  found  from  the  series 
by  this  modulus,  we  will  have  a  table  of  common  logarithms. 


ADA^ANTAGES  OF  THE  COMMON  SYSTEM. 

422.   1.  A  fixed  law  for  the  characteristic. 

The  characteristic  of  a  logarithm  is  its  entire  part.  Thus,  2  is  the 
characteristic  in  the  logarithm  2-302585093. 

Whatever  may  be  the  base  of  the  system,  the  logarithms  of  the 
powers  of  that  base  will  contain  no  decimals,  and,  therefore,  be  charac- 
teristics only.  Thus,  let  7  be  the  base,  then  the  log.  7  —  1, 
log.  49  =  2,  &c. 

In  the  common  system,  with  the  base  10,  we  have  log.  10  =  1, 
log.  100  =  2,  log.  1000  =  3,  &c.,  and  we  see  that  the  characteristic  is 
always  one  less  than  the  number  of  figure  places  in  the  number.  This 
is  true,  whether  the  number  be  an  exact  power  of  the  base  or  not ; 
thus,  512,  lying  between  100,  whose  logarithm  is  2,  and  1000,  whose 
logarithm  is  3,  has  for  its  logarithm  2  and  a  certain  decimal.  Now, 
10  is  the  only  number  in  the  whole  range  of  numbers,  which,  taken  as 
a  base,  will  give  logarithms  whose  characteristics  are  formed  according 
to  a  fixed  law. 

In  a  table  of  common  logarithms,  the  characteristics  are  not  written, 
since  they  are  always  one  less  than  the  mimher  of  figures  in  the  given 
numbers. 

2.  A  tahle  of  logarithms  for  decimals  need  not  be  constructed. 

Since,  Log.  -1  =  log.*  -J^  =  log.  1  —  log.  10  =  —  1. 
Log.  -01  =  log.  jJ  0  =  ^og-  1  —  'og.  100  =  —  2. 
Log.  -001  =  log.  j-^oT)  =  log-  1  —  log.  1000  =  —  3. 


LOGARITHMS.  405 

And,  since,  in  general,  log.  :=  log.  1  —  m  log.  10  =  —  m,  it  is 

plain  tliat  the  characteristic  of  a  decimal  is  negative,  and  one  greater 
than  the  number  of  ciphers  between  the  decimal  point  and  the  first 
significant  figure.  This  is  true,  also,  when  the  decimal  is  not  the  re- 
ciprocal of  some  power  of  the  base.  Thus,  log.  -08  =  log.  -f^  =  log. 
8  —  log.  10  =  a  decimal  —  unity.  Hence,  the  log.  -08  must  have  a 
negative  characteristic  equal  to  minus  unity.  Moreover,  if  we  have  a 
decimal,  such  as  -00278  =  jxhhj1)0'  ^^  ^^'^  ^^^'^  'og-  '00278  =  log. 
278  —  log.  100000.  And,  since  log.  278  is  2  whole  number,  plus  a  cer- 
tain decimal;  and,  since  the  log.  100000  is  5,  the  log  -00278  =  —  3, 
whole  number  plus  a  certain  decimal.  We  see,  from  this  example,  that  the 
logarithm  of  a  decimal  is  obtained  by  regarding  the  decimal  as  a  whole 
number,  and  prefixing  a  negative  characteristic,  one  greater  than  the  num- 
ber of  ciphers,  between  the  decimal  point  and  the  first  significant  figure. 
The  foregoing  reasoning  can  be  extended  to  any  decimal  whatever,  be- 
cause, change  the  decimal  into  an  equivalent  vulgar  fraction,  it  will  be 
seen  that  the  number  of  figure  places  in  the  denominator  exceed  those 
in  the  numerator  by  one  more  than  the  number  of  ciphers  between  the  de- 
cimal point  and  the  first  significant  figure  of  the  given  decimal.  Now,  if 
the  base  were  any  other  number  whatever  than  10,  it  is  plain  that  a  table 
of  logarithms  for  whole  numbers  would  not  be  applicable  to  decimals. 

3.  The  logarithm  of  a  mixed  decimal,  such  as  42-733,  can  he  found 
from  a  table  of  logarithms  constructed  to  the  base  10,  b^  regarding  the 
mixed  number  as  a  whole  number,  and  prefixing  to  the  decimal  p«r< 
(f  the  logarithm  found  in  the  table  a  characteristic  one  less  than  the 
number  of  figure  places  in  the  entire  p)art  of  the  mixed  number. 

For,  log.  42-733  =  log.  *f^^\^  =  log.  42733  —  log.  1000  =  4  + 
a  certain  decimal  —  3  =  1  +  a  decimal. 

The  foregoing  rule  for  finding  the  logarithm  of  a  mixed  number  can 
evidently  be  applied  to  any  mixed  number  whatever. 

4.  The  logarithms  of  numbers  which  difi"er  only  in  the  number  of 
annexed  ciphers,  will  differ  only  in  their  characteristics. 

Thus,  the  logarithms  of  125,  1250,  12500,  and  125000,  difi'er  only 
in  their  characteristics.  For,  log.  1250  =  log.  10  -f  log.  125  =  1  -f- 
log.  125 ;  log.  12500  =  log.  100  -f  log.  125  =  2 -|-log.  125 ;  log.  125000 
=  log.  1000  +  log.  125  =  3  -f  log.  125. 

This  property  of  common  logarithms  is  very  important,  since  it  saves 
the  trouble  (as  will  be  seen  more  fully  hereafter)  of  constructing  a  table 
for  numbers  whose  figure  places  exceed  four. 


406 


T.  O  (J  A  U  1  T  II  M  S  . 


APPLICATION  OF  COMMON  LOGAPJTIIMS. 

423.  The  base  of  the  coiniuon  system  being  10,  the  characteristic  or 
entire  part  of  the  logarithms  need  not  be  written  iu  the  tables,  since,  as 
we  have  seen,  it  is  always  one  less  than  the  number  of  figure  places  iu 
the  given  number.  The  decimal  parts  of  the  logarithms  are  then  only 
written  in  the  tables,  and  are  generally  carried  as  far  as  six  places. 
The  first  vertical  column  on  the  left  in  the  tables,  contains  the  given 
numbers ;  the  last  vertical  column,  marked  D,  contains  the  tabular  dif- 
ference corresponding  to  two  given  logarithms,  and  is  constructed  by  a 
method  hereafter  to  be  explained.  The  intermediate  vertical  columns, 
marked  at  top,    0,  1,  2,  3,  &c.,  contain  the  decimal  parts  of  logarithms. 

The  following  table  exhibits  the  logarithms  and  numbers  between  100 
and  130.      When  the  given   number  contains  four  figure  places,  the 


000000 
•4321 
8P«0 

01 2837 
7033 

"5306 
9384 

033424 
7426 

041393 
5323 
9218 

053078 
6905 

0GO698 
'4458 
SI  86 

071882 
5547 

079181 

082785 
0360 
9905 

093422 
6910 

100371 
3804 
7210 

110590 

113043 


■3S0 
4230 
8046 
1S29 
5580 
9298 
2985 
6640 

•206 
3861 
7426 

•963 
4471 
7951 
1403 


7071 
•706 
4451 
8094 
1707 
5291 
S845 

5806 
9335 
2777 
6191 
9579 
2940 
6276 


first  three  will  be  found  in  the  first  vertical  column  on  the  left,  and  the 
fourth  must  be  sought  at  the  top  of  the  page,  among  the  numbers  be- 
tween 0  and  9,  inclusive.  Thus,  121  may  be  found  in  the  first  vertical 
column  on  the  left,  and  the  last  figure  of  1215  will  be  found  in  the 
column  marked  5  at  top.  The  logarithm  of  121  is  written  just  oppo- 
site to  it  iu  the  column  marked  0;  1210  will  obviously  have  the  same 


LOGARITHMS.  407 

decimal  part  of  its  logaritlim,  but  a  characteristic  greater  by  unity ;  the 
logarithm  of  1215,  on  the  other  hand,  must  not  only  exceed  the  loga- 
rithm of  121  in  characteristic,  but  also  in  decimal  part,  and  its  decimal 
part  is  then  taken  from  the  column  marked  5. 

TO  FIND  THE  LOGARITHM  OF  A  GIVEN  NUMBER  BELOW  10,000. 

424.  When  the  number  is  less  than  100,  the  decimal  part  of  the 
logaritlim  is  found  in  the  table  just  opposite  the  given  number.  The 
characteristic  will  be  determined  by  the  number  of  figure  places  in  the 
number.  For  numbers  above  100  and  below  10,000,  we  have  the 
following 

RULE. 

Loohfor  the  first  three  figures  of  the  given  number  in  the  column 
marked  N,  and  if  it  contains  hut  three,  the  decimal  part  of  its  loga- 
rithm icill  he  found  Just  ojijmsite  these  three  figures  in  the  column 
marked  0.  But  if  there  are  only  four  figures  in  this  column,  opposite 
the  given  numher,  the  two  left  hand  figures  above,  in  the  same  column, 
where  six  figures  aretcritten,  must  he  prefixed  to  the  four  figures  found. 
Thus,  the  decimal  part  of  the  logarithm  of  110,  is  041398,  and  the 
entire  logarithm,  20i1SdS.  The  decimal  2mrt  of  the  logarithm  of 
112  is  Oid-218,  the  first  tu-o  figures  being  found  opposite  110.  The 
entire  logarithm  is  then  2-049218.  The  logarithm  of  117  is  2-06816, 
and  of  120,  2079181. 

425.  When  the  given  number  contains  four  figures,  look  for  the  first 
three  in  the  column  marked  N.  The  last  four  of  the  decimal  part  of 
its  logarithm  will  be  found  just  opposite,  in  the  column  marked  at  top 
by  the  fourth  figure.  To  these  four  figures  must  be  prefixed  two  deci- 
mals in  the  column  marked  zero,  either  found  opposite  the  first  three 
figures,  or  above,  where  six  figures  occur.  Thus,  to  get  the  logarithm 
of  1213,  we  first  find  121  in  the  first  column,  and  then  look  just  oppo- 
site, in  the  column  marked  3,  where  we  find  3801,  and  prefix  to  it  08, 
taken  from  the  column  marked  0.  Hence,  the  logarithm  of  1213  is 
8-083861.  The  logarithm  of  1283  is  3-108227,  the  two  left  hand 
figures  of  its  decimal  part  being  found  above,  where  six  figures  occur. 

When,  however,  we  pass  over  a  decimal  with  a  point  prefixed,  the 
first  two  figures  are  to  be  found  below,  and  not  above.  Thus,  the  loga- 
rithm of  1234  is  3-091315.  In  like  manner,  the  logarithm  of  1231 
is  3-090258.     The  place  of  the  point  is  always  supplied  by  zero.     If 


408  LOGARITHMS. 

the  left  hand  figures  were  not  taken  below,  the  logarithm  of  1231,  1232, 
&c.,  would  be  less  than  the  logarithm  of  1230,  1229,  &c. 

EXAMPLES. 

1.  Required  the  logarithm  of  129.  Ans.  2-110590. 

2.  Required  the  logarithm  of  1290.  Ajis.  3410590. 

3.  Required  the  logarithm  of  1297.  A^is.  3-112940. 

4.  Required  the  logarithm  of  1285.  Ans.  3-108903. 

5.  Required  the  logarithm  of  1233.  A71S.  3-090963. 

6.  Required  the  logarithm  of  1236.  Ans.  3-092018. 

7.  Required  the  logarithm  of  119.  Ans.  2-075547. 

8.  Required  the  logarithm  of  1191.  A71S.  3-075912. 

TO  FIND  THE  LOGAPJTIOI  OF  A  NUMBER  ABOVE  10,000. 

426.  Cut  off  on  the  right  all  the  figures,  except  the  first  four  of  the 
highest  denomination.  Find  the  decimal  part  of  these  four  left  hand 
figures,  as  before,  and  prefix  a  characteristic,  one  less  than  the  number  of 
figure  places  in  the  given  number  previous  to  cutting  off.  jNIultiply 
the  tabular  difference  by  the  figures  cut  off  on  the  right,  and  from  this 
product  cut  off  as  many  places  for  decimals  as  there  are  figure  places  in 
the  multiplier.  Add  the  result  to  the  logarithm  before  found.  Thus, 
to  find  the  logarithm  of  129451,  we  cut  off  51,  and  find  the  decimal 
part  of  1294  to  be  111934.  To  this  we  prefix  a  characteristic,  5,  and 
we  have  5-111934,  which  is  truly  the  logarithm  of  129400  (Art.  422). 
Next,  we  multiply  the  tabular  difference,  335,  by  51,  and  cut  off  85 
from  the  product,  17085 ;  we  would  then  have  170  to  add  to  5-111934 ; 
but,  as  -85  is  greater  than  |,  we  increase  170  by  1,  because  171  is 
nearer  to  170-85  than  is  170.  Adding  then  171,  we  have  5-112105 
for  the  logarithm  of  129451.  In  general,  whenever  the  decimal  cut 
off'  in  the  product,  which  results  from  multiplying  the  tabular  difference 
by  the  right  hand  figures  of  the  given  number,  exceeds  ^,  the  last 
figure  of  the  entire  part  of  the  product  must  be  increased  by  1. 

The  explanation  of  the  process  is  simple.  The  tabular  diffe- 
rence expresses  the  difference  between  the  logarithms  of  consecutive 


LOGARITHMS.  409 

numbers.  Thus,  the  logarithm  of  1293  is  3-1599 ;  and  of  1294  is 
3-1934,  and  their  difference,  335,  is  written  in  the  column  marked  D. 
This  column,  it  is  plain,  will  moreover  express  the  difference  between 
the  logarithms  of  numbers  which  differ  by  10,  100,  1000,  &c.  "We 
have  seen  that  the  logarithm  of  129400  was  5-11934,  but  we  were 
required  to  find  the  logarithm  of  129451,  and  it  remains  to  be  seen  how 
much  the  logarithm  of  129400  is  to  be  increased.  We  employ  the 
principle,  that  the  difference  between  any  two  numbers  is  to  the  difference 
of  their  logarithms,  as  the  difference  between  any  two  other  numbers  is 
to  their  difference  of  logarithm.  Hence,  129400  — 129300  :  335  :  : 
129451  — 129400  :  x,  or  100  :  335  :  :  51  :  z,  in  which  x  represents 
the  augment  to  the  logarithm  of  129400,  to  give  the  logarithm  of 


EXAMPLES. 

1.  Required  the  logarithm  of  1309958.  Ans.  6-117259. 

In  this  example  three  places  arc  cut  off  for  decimals,  because  we  took 
the  difference  between  1309000  and  1308000.  The  first  term  of  the 
proportion  was  then  1000. 

2.  Required  the  logarithm  of  13080000.  Ans.  7-116608. 

3.  Required  the  logarithm  of  13087654.  Aiis.  7-116862. 

TO   FIND    A    NUMBER   CORRESPONDING    TO    A    GIVEN 
LOGARITHM. 

427.  When  the  decimal  part  of  the  logarithm  can  be  found  in  the 
tables,  the  first  three  places  of  the  required  number  will  be  found  on 
the  left,  immediately  opposite  this  logarithm,  and  in  the  column  marked 
N.  The  fourth  figure,  which  is  to  be  annexed  to  the  three  already  found, 
is  to  be  taken  from  the  number  of  the  column  in  which  the  decimal 
part  of  the  logarithm  was  written.  Then  point  off,  from  the  left  of  the 
number  thus  obtained,  one  more  place  for  the  entire  part,  than  is  indi- 
cated by  the  number  of  units  in  the  characteristic.  Thus,  the  number 
corresponding  to  the  logarithm  2-107888  is  128-2,  and  to  1-107888  is 
12-82.  So,  the  number  corresponding  to  the  logarithm  3-088136  is 
1225,  and  contains  no  decimal  part. 

But,  when  the  decimal  part  cannot  be  exactly  found,  take  the  next 
less  logarithm  from  the  tables,  and  subtract  this  from  the  given  loga- 
35 


410  LOGARITHMS. 

ritlim.  Annex  two  or  more  ciphers  to  this  difference,  and  divide  by 
tlic  tabular  difference.  Annex  the  quotient  to  the  number  correspond- 
ing to  the  next  less  logarithm,  and  you  will  have  the  required  number. 
Thus,  let  it  be  required  to  find  the  number  corresponding  tS  the  lo- 
garithm 4-1132800.  The  next  less  logarithm  is  4-1132750,  and  the 
corresponding  number  12980 ;  annexing  4  ciphers  to  50,  the  difference 
of  logarithms,  and  dividing  by  335,  the  tabular  difference,  we  have  a 
<luotient,  149,  which  is  to  be  annexed  as  a  decimal  to  12980.  Hence, 
the  number  sought  is  12980-149.  It  is  plain  that  we  may  annex  as 
many  ciphers  as  we  please  to  the  difference  of  logarithms.  The  pro- 
cess just  described  being  the  reverse  of  the  preceding,  requires  no  ex- 
planation. 

EXAMPLES. 

1.   Required  the  number  whose  logarithm  is  4-07850. 


2.   Required  the  number  whose  logarithm  is  7-116608. 


11981-18. 

13080000. 


U.  Required  the  number  whose  logarithm  is  8-075550. 

Ans.  119000243-85. 

4.  Required  the  number  whose  logarithm  is  3-017249. 

Ans.  1040-52. 

5.  Required  the  number  whose  logarithm  is  3-153209. 

A71S.  1423-013. 


Remarks. 

When  only  one  cipher  has  to  be  annexed  to  the  difference  between 
the  given  and  next  less  logarithm,  to  make  it  divisible  by  the  tabular 
difference,  no  cipher  is  prefixed  to  the  quotient.  But  when,  (as  in 
example  5),  two  ciphers  must  be  annexed  to  make  the  division  possible, 
then  one  cipher  must  be  prefixed  to  the  quotient.  And  when,  (as  in 
example  3),  three  ciphers  must  be  annexed,  two  must  be  prefixed  to  the 
quotient.  And,  in  general,  the  number  of  ciphers  prefixed  is  one  less 
than  the  number  of  ciphers  annexed,  to  make  the  division  possible. 


k;. 


LOGARITHMS.  V       .  -iVl 


GENERAL    EXAMPLES. 

1.  Piequired  the  product  of  1-040,  by  1-055  and  10*77. 

Ans.  11-81687. 
For  log.  1-040  =  0-017033 

log.  1-055  =  0023252 

log.  10-77  =  1-032216 
And  logarithm        1-072501  corresponds  to  11-81687. 

2.  Eequired  the  quotient  of  10-22  by  1016.  Ans.  10-059. 

For  log.  10-22  =  1-009451 
and    \o<r.    1-016  =  0006894 


and  their  difference,  1-002557,  corresponds  to  10-059. 

8.  Raise  11-48  to  the  third  power.  Ans.  1512-95. 

For  log.  11-48  =  1  059942 
3 


3-179826  =  log.  1512-95. 

4.  Required  the  thirty-second  root  of  1172.  Ans.  1-21871. 

loo-.  1172       3  06892S 

For  we  have     ".,.       =  ^r^  =  -095904. 

o2  o2 

And  the  corresponding  number  is  1-21871. 


SOLUTION  OF  EXPONENTIAL  EQUATIONS  BY  MEANS  OF 
LOGARITHMS. 

428.  Suppose  we  have  the  exponential  equal,  a^  =i'j  then,  x  log. 

a  =  I02;.  p,  and  .v  =  ~ —  ;  that  is,  the  root  is  equal  to  the  logarithm 
^  ^  '  log.  a  '  ^  ° 

of  p  divided  by  the  logarithm  of  a,  and  this  root  itself  may  be  found 
by  means  of  logarithms  for  loc.  .r  =  I02;.  ,  ''^"  =  losr.  losr.  p  —  lopr. 
log.  a. 

EXAMPLES. 

1.  Required  the  value  of  .x  in  the  equation,  10^  =  10. 

Ans.  X  =  1. 
For,  X  log.  10  =  log.  10,  or  x  (!)  =  1,  or  x  =  1. 


412  ARITHMETICAL    PROGRESSION. 

2.  Required  the  value  of  x  in  the  equation,  12^  =  13. 

Ans.  1-032. 

For,     k)g.  13  =  1-113943     I     log.  1-113943  :==  0-046862 
and,         log.  12  =  1-079181     |     log.  1-079181  :=  0  033094 

0-013768. 
Hence,  log.  x  =  0-013768,  or  x  =  1-032. 

3.  Required  the  value  of  x  in  the  equation,  lOO''  =  1000. 

Ans.  ic  =  |. 

4.  Required  the  value  of  x  in  the  equation,, 10^  =  125. 

A71S.  X  =  2-090910. 

5.  Required  the  value  of  x  in  the  equation,  100^  =  125. 

Ans.  X  =  1-048455. 

6.  Required  the  value  of  x.  in  the  equation,  11*  =  11. 

Ans.  a:  =  1. 

For,  log.  X  =  log.  log.  11  —  log.  log.  11  =  log.  1-041393  —  log. 
1-041393  ^  0-017614  —  0-017614  =  0.  Now,  since  log.  x  =  ^,x 
must  be  unity. 

The  result  in  this  case  might  have  been  anticipated,  but  we  have 
carried  out  the  process  to  show  its  truthfulness. 


ARITHMETICAL   PROGRESSION. 

429.  Quantities,  which  increase  or  decrease  by  a  common  diffe- 
rence, are  said  to  be  in  Arithmetical  Progression. 

Thus,  2,  4,  6,  8,  10,  &c.,  constitute  an  arithmetical  progression,  the 
common  difference  being  2.  In  like  manner,  10,  8,  6,  4,  2  constitute, 
also,  an  arithmetical  progression,  the  common  difference  being  2. 
When,  as  in  the  first  case,  the  common  difference  is  an  increment,  the 
series  is  called  ascending ;  and  when,  as  in  the  second  case,  the  com- 
mon difference  is  a  decrement,  the  series  is  called  descending.  The 
quantities,  a,  a  -\-  d,  a  -\-  2d,  a  +  3c7,  &c.,  constitute  an  ascending 
series;  and  a,  a  —  d,  a  —  2(7,  a  —  oc7,  constitute  a  descending  series. 


ARITHxMETICAL    PROGRESSION.  413 

It  is  evident,  likewise,  that  the  quantities,  a  +  3(7,  a  -\-  2d,  a  +  d,  a, 
form  a  descending  series ;  and,  in  general,  an  ascending  series,  when 
read  in  reverse  order,  becomes  a  descending  series,  and  conversely. 

It  is  plain  that  the  natural  numbers,  1,  2,  3,  4,  5,  6,  7,  &c.,  are  in 
arithmetical  progression,  the  common  diiference  being  1. 

It  is  plain  that,  if  we  know  any  term  of  an  ascending  series,  we  can 
find  the  succeeding  term  by  adding  to  the  known  term  the  common 
difference.  And,  if  we  wish  to  find  the  term  succeeding  any  known 
term  of  a  decreasing  series,  we  have  only  to  subtract  the  common  diffe- 
rence from  the  known  term.  In  this  manner,  we  may  find  any  term 
of  a  series  by  a  continued  addition  or  subtraction  of  the  common 
difference.  But,  as  in  a  long  series,  this  process  would  be  tedious, 
it  becomes  necessary  to  deduce  a  formula  by  which  any  term,  as 
the  71*^,  may  be  found  without  going  through  a  tedious  addition  or 
subtraction. 

FORMULA  FOR  THE  Kth  TERM. 

430.  The  second  term  of  the  ascending  series,  a,  a  -\-  d,  a  +  2d, 
&c.,  is  a  +  d ;  that  is,  the  common  difference  is  added  to  the  first  term 
one  less  number  of  times  than  the  place  of  the  term.  The  third  term 
is  a  -f  2d;  that  is,  the  first  term  is  increased  by  the  common  difference, 
taken  once  less  than  the  place  of  the  term.  We  discover  the  same  law 
of  formation  in  regard  to  the  fourth  term,  and  all  succeeding  terms. 
Hence,  calling,  the  n^^  term  /,  wc  have  /  =  a  -f  (ii  —  l)d ;  that  is, 
the  n*"*  term  is  equal  to  the  first  term,  increased  hy  the  common  diffe- 
rence, taken  once  less  than  the  jyJace  of  the  term. 

If  the  series  is  decreasing,  d  is  negative,  and  the  formula  becomes 
7  =  a  —  (n  —  V)d.  The  first  formula  includes  the  second  ;  and  the 
first,  therefore,  is  only  necessary,  care  being  taken  to  attribute  to  d  its 
proper  sign. 

For  an  ascending  series  the  first  term  is  always  the  least,  and  for  a 
descending  series  the  first  term  is  always  the  greatest. 

The  foregoing  formula,  l=za  -\-  (n  —  V)d,  may  be  used  to  obtain 
the  last  term,  for  n  may  represent  any  term  whatever.  When  the  com- 
mon difference  is  equal  to  the  first  term,  that  is,  d  =  a,  then,  l=za 
-f  na  —  a  =  na.  The  n^^  term  is  then  equal  to  the  first  term  into 
the  number  of  terms.  When  the  number  of  terms  is  unity,  that  is, 
n  =  1,  then.  l=za,  as  it  ought  to  be. 
35*' 


414  ARITHMETICAL    PR  OG  11  E  S  SIO  N  . 


EXAMPLES. 

1.  The  first  term  of  an  ascending  series  is  2,  the  common  difference 
3,  required  the  fifth  term.  Ans.  14. 

2.  Required  the  tenth  term  of  the  same  series.  Ans.  29. 

8.  Required  the  last  term  of  an  ascending  series,  of  which  the  first 
term  is  10,  the  common  difference  5,  and  the  number  of  terms  11. 

A71S.  GO. 

4.  Required  the  eleventh  term  of  a  descending  series,  of  which  the 
first  term  is  GO,  and  the  common  difference  5.  Atis.  10. 

5.  Required  the  tenth  term  of  an  ascending  series,  of  which  the 
first  term  is  10,  and  the  common  difference  10.  Ans.  100. 

G.  Required  the  last  term  of  a  decreasing  series,  of  which  the  first 
term  is  100,  the  common  difference  10,  and  the  number  of  terms  10. 

Ans.  10. 

FORMULA  FOR  THE  FIRST  TERM. 

431.  The  formula,  I  =  a  +  (n  —  V)d,  contains  four  quantities,  and, 
of  course,  if  any  three  are  given,  the  fourth  can  be  determined.  By 
transposition  and  reduction,  we  have  a  =  1  —  (?i  —  V)d.  That  is,  the 
first  term  is  equal  to  the  last,  or  n*"*  tet-m,  m  inns  the  number  of  terms, 
less  one,  'into  the  common  difference. 

The  formula  is  applicable  both  to  an  ascending  and  to  a  descending 
series,  by  attributing  the  proper  sign  to  d. 

EXAMPLES. 

1.  The  fifth  term  of  an  ascending  series  is  14,  and  the  common 
difference  3.     Required  the  first  term.  Ans.  2. 

2.  The  tenth  term  of  the  same  series  is  29.    Required  the  first  term. 

Ans.  2. 

3.  Required  the  first  term  of  an  ascending  series,  whose  last  term  is 
60,  common  difference  5,  and  the  number  of  terms  11.         Ans.  10. 

4.  Required  the  first  term  of  a  descending  series,  of  which  the  last 
term  is  10,  the  common  difference  5,  and  the  number  of  terms  11. 

Ans.  60. 


ARITHMETICAL    PROGRESSION.  415 


FORMULA  FOE  THE  COMMON  DIFFERENCE. 
432.  From  the  formula,  I  =  a  +  (n  —  l)c7,  we  get  d  = ,  that 

is,  the  common  difference  is  equal  to  the  last  term,  viinus  the  first  term, 
divided  hy  the  number  of  terms  less  one. 

For  a  descending  series,  d  is  negative,  and,  by  multiplying  both 

members  by  minus  unity,  we  get  d  =  -,  that  is,  the  common  dif- 
ference of  a  descending  series  'is  equal  to  the  first  term,  minus  the  last 
term,  divided  hy  the  number  of  terms  less  one  When  I  =z  a,  d  will  be 
zero,  as  it  ought  to  be.  When  n  =  1,  d  =  cc.  The  symbol  of  absurdity 
ought  to  appear  under  the  hypothesis  n  :=  1 ;  for,  when  there  is  but 
one  term,  there  can  be  no  common  diiFerencc,  and  therefore  the  assump- 
tion of  its  existence  is  absurd. 


1.  The  last  term  is  50,  the  first  term  10,  and  the  number  of  terms  8. 
Required  the  common  difference.  Ans.  20. 

2.  The  last  term  is  10,  the  first  term  50,  and  the  number  of  terms  3. 
Required  the  common  difference.  Ans.  20. 

3.  The  last  term  is  12,  the  first  term  1,  and  the  number  of  terms  4. 
Required  the  common  difference.  Ans.    y . 

FORMULA  FOR  THE  NUMBER  OF  TERMS. 

433.  From  I  =  a  -\-  (n  —  l)d,  we  get  n  =  1  -| — ,  that  is,  the 

number  of  terms  is  equal  to  unity,  added  to  the  quotient  arising  from 
dividing  the  difference  between  the  first  and  last  terms  by  the  common 
difference. 

The  formula  is  applicable  to  a  decreasing  series,  by  attributing  the 
appropriate  sign  to  d;  when  I  =  a,  n  will  be  equal  to  1,  when  d  =  0, 
n  =  cc  . 

EXAMPLES. 

1.  The  last  term  is  50,  the  first  term  10,  and  the  common  difference 
20.     Required  the  number  of  terms.  Ans.  3. 


410  AIUTIIMKTICAL     PROURESSION. 

2.  The  kst  term  is  12,  the  first  term  1,  and  the  common  difference 
y.     Required  the  number  of  terms.  Ans.  4. 

3.  The  last  term  is  50,  the  first  term  10,  and  the  common  difference 
1.     Required  the  number  of  terms.  Ans.  41. 


FORMULA  FOR  THE  SUM  OF  THE  TERMS. 

434.  We  might  get  the  sum  of  the  terms  by  actually  performing  the 
addition ;  but,  when  the  series  contains  many  terms,  this  operation  would 
be  difficult,  and  we  are  enabled  to  deduce  a  formula  which  abridges  the 
work.  This  formula  is  deduced  from  the  remarkable  property  of  an 
arithmetical  progression,  that  the  sum  of  any  two  terms  at  equal  distance 
from  the  two  extremes  is  equal  to  the  sum  of  the  two  extremes.  To 
show  this,  any  term,  as  the  m^^,  counting  from  the  first  term,  will  be 
expressed  by  a  -|-  (m  —  l)c7,  and  the  m^^  term,  counting  from  the  right, 
will  be  expressed  by  I — (m  —  l)d;  and  the  sum  of  these  terms  is 
plainly  a  +  I,  a  denoting  the  first  term  and  I  the  last  term. 

Calling  S  the  sum  of  the  terms,  we  have  S'=a+ (a  + <:?)  +  («  + 2(7) 
+  (a  +  3(7)  -\-  (a  +  4d)  ....  to  7,  and  reversing  the  series,  S  — 
/-j-(I — d)  +  (I — 2(7)  +  (7  —  3(7)  .  .  .  to  rt.  Adding  these  equations 
member  by  member,  we  get  2S  =  (a  +  7)  +  (a  +  7)  +  (a  +  7)  4- 
&c.,  up  to  71  terms. 

Hence,  2s  =  (a  +  7)«,  or  s  =  ^^ — ,   that   is,    the   sum   of  the 

terms  is  equal  to  the  half  sum  of  the  first  and  last  terms,  into  the  num- 
her  of  terms. 

When  11  =  2,  s  =  a  -{-I,  as  it  ought  to  be.  When  7  =  a,  s  =  na, 
the  common  difference  is  then  zero.  When  ?i=  4,  s  ^  2  (a  -|-  I),  as  it 
ought  to  be.     When  ?i  =  0,  s  =  0,  as  it  ought  to  be. 

The  sum  of  the  terms  can  be  determined  when  a,  I  and  n  are  known, 
or  can  be  found  from  the  data. 


1.  Find  the  sum  of  the  natural  numbers,  1,  2,  3,  4,  5,  6,  7,  8,  9. 

Ans.  45. 

2.  The  first  term   of  a  series  is  10,  the   last  term  100,  and  the 
number  of  terms  5.     Required  the  sum  of  the  terms. 

Ans.  275. 


ARITHMETICAL    PROGRESSION.  417 

3.  The  first  term  of  an  ascending  series  is  5,  the  common  difference 
5,  and  the  number  of  terms  5.     Required  the  sum  of  the  terms. 

A71S.  75. 

4.  The  last  term  of  an  ascending  series  is  60,  the  number  of  terms 
4,  and  the  common  difference  2.     Required  the  sum  of  the  terms. 

Ans.  228. 

5.  The  last  term  of  a  descending  series  is  60,  the  number  of  terms 
4,  and  the  common  difference  2.     Required  the  sum  of  the  terms. 

Ans.  252. 

6.  The  last  term  of  an  ascending  series  is  50,  the  first  term  6,  and 
the  common  difference  2.     Required  the  sum  of  the  terms. 

Ans.  644. 

7.  Find  s,  when  a,  d,  and  n,  are  known. 

Ans.  S  =  i\2a  +  (n  —  l)d\n. 

8.  The  first  term  of  an  ascending  series  is  1,  the  common  differ- 
ence 1,  and  the  number  of  terms  10.  Required  the  sum  of  the 
terms.  Ans.  55. 

9.  How  many  strokes  does  the  common  clock  strike  in  12  hours  ? 

Ans.  78. 

10.  A  body  falling  in  vacuo  will  pass  over  about  16  feet  the  first 
second,  48  the  next  second,  80  the  third  second,  &c.  How  far  will 
it  fall  in  20  seconds,  and  what  space  will  it  pass  over  in  the  last  second  ? 

Ans.  Entire  space,  6400  feet;  in  last  second,  624  feet. 

11.  A  traveller  goes  5  miles  the  first  day,  15  the  second,  25  the 
third,  &c.  Required  the  space  that  will  have  been  passed  over  at  the 
end  of  the  tenth  day.  Ayis.  500  miles. 

12.  Find  the  sum  of  the  natural  numbers,  1,  2,  3,  &c.,  up  to  n  terms. 

-4ns.   i  n(n  -i-  1). 

13.  Find  the  sum  of  the  odd  numbers,  1,  3,  5,  7,  &c.,  up  to  n  terms 

Ans.    n^. 

14.  Find  the  sum  of  the  even  numbers,  2,  4,  6,  8,  10,  &e.,  up  to  n 
terms.  Ans.   n  (n  +  1). 

15.  Find  the  sum  of  the  numbers,  6,  12, 18,  24,  &c.,  up  to  w  terms. 

Ans.  3n  (n  +  1). 


418  ARITHMETICAL    rROGRESSION. 

16.  Find  the  sum  of  the  odd  uumbers,  1,  2,  8,  &c.,  up  to  73. 

Ans.  5329. 

17.  Find  the  sum  of  the  even  numbers,  2,  4,  6,  &c.,  up  to  twenty- 
five  terms.  Ans.  650. 

18.  Find  the  sum  of  the  numbers,  6,  12,  18,  &c.,  up  to  the  num- 
ber 72.  Ans.  468. 

19.  Find  s,  when  I,  d  and  n,  are  known. 

Ans.  ^  =  l\2l—{n  —  V)d\n. 


TO  FIND  THE  ARITHMETICAL  MEAN. 

435.  The  arithmetical  mean  between  several  quantities,  is  the  quo- 
tient arising  from  dividing  their  sum  by  their  number.  The  arithmeti- 
cal mean  between  two  quantities,  m  and  w,  is  half  their  sum,  that  is, 
i  (m  +  n).  We  have  seen  that  a  geometric  mean  between  two  quantities 
is  the  square  root  of  their  product. 

From  the  definition,  the  arithmetical  mean  between  any  number  of 

terms  in  progression  must  be  —  = ^     =  J  (a  -j-  ?),  that  is,  the 


arithmetical  mean  is  equal  to  the  half  sum  of  the  extremes. 


Corollary. 

The  first  and  last  terms  may  be  found  when  the  arithmetical  mean, 
the  number  of  terms,  and  the  common  difi"erence,  are  known. 

For,  from  S  =  ^^ ,  we  get  a  = (a  +  {n  —  1)  d),    or 

a  =  2^l  —  a  —  {n  —  r)d.     Hence,  a  =  M  —  ^^^^=^. 

And  since  Z  =  a  +  {n  —  V)d,  we  get  Z  =  M  +  ^^ 

The  employment   of  the   arithmetical   mean   frequently  facilitates 
the  solution  of  a  problem  in  progression. 


ARITHMETICAL    PROGRESSION.  419 


EXAMPLES. 

1.  The  sum  of  four  numbers  in  arithmetical  progression  is  20,  and 
their  continued  product,  384.     What  are  the  numbers? 

Ans.  2,  4,  6,  8. 


For, 

M  = 

^=='°  = 

5.     And  the  first  term,  a  =  M  —  - 

11  1.  )  Ur 

2 

=  5- 

-1^; 

the  second 

term,  5  —  M; 

the  third  term,  5  + 

Id;  the 

fourth 

term. 

5  +  id. 

Hence,  by  the 

conditions  of  the 

problem, 

(5-|cZ)(5 

—  id)  (5  +  id)  (5  +  |(7) 

=  384, 

or  625  — 

250^^ 
4      "'' 

9d* 
16  = 

384. 

Making  d^ 

•  =  2^,  we  . 

^et,  after  reduction,  ^  — 

lOOOy 
9 

38r)6 
a   ■ 

Hence, 

.y  = 

500       464 
9    "^    9 

964 

=  +  4,or+— . 

The  last 

value  being  rejected, 

we  have  d^=  +  i.  Hence,  d  =  ziz2.  The  positive  value  will  give 
a  =  5  —  |fZ  =  2;  and  the  second  term,  4;  the  third,  6;  the  fourth,  8. 
The  negative  value  of  d  will  give  the  series  in  reverse  order,  and  we 
see  here  a  change  of  sign  followed  by  a  change  of  direction. 

The  same  problem  might  be  solved  by  making  2x  the  common  dif- 
ference, and  7/  —  Sx,  the  smaller  extreme.  Then  the  4  terms  will  be 
represented  by  y  —  3a;,  y  —  x,  y  -\-  x,  and  y  +  Sx.  And  we  have  the 
two  equations,  ^  —  3x+y  —  x  +  1/  +  x  +3/+3x  =  4y  =  20,  and 
(ij  —  Sx)  (y  +  Sx)  {9/  —  X)  (y  +  x)  =  (f  —  9x')  (/_^2)  =y_ 
lOa^y  +  9x*  =  384.  Substituting,  in  this  last  equation,  the  value  of 
y  drawn  from  the  first,  we  get  625  —  250x2  +  9a;'  =  384.  From 
which,  x=  zizl.     Then  y  —  3x  =  5  —  3  =  2,  and  y  —  x  =  4,  &c. 

The  student  will  readily  perceive  the  advantage  of  representing  the 
first  extreme  by  y  —  Sx,  and  the  common  difibrence  by  2x. 

2.  The  sum  of  5  numbers  in  arithmetical  progression  is  15,  and  their 
product  120.     What  are  the  numbers  ?  Ans.  1,  2,  3,  4,  5. 


420  ARITHMETICAL    PROGRESSION, 


INSERTION  OF  MEANS  BETWEEN  THE  EXTREMES  OF  A 
PROPORTION. 

436.  The  formula,  d  =  -,  enables  us  to  insert  any  number  of 

^  n  —  1'  •' 

means,  when  the  extremes  are  known.  Let  it  be  required  to  insert  3 
means  between  1  and  9.  Then  a  :=  1,  and  Z  =  9,  and,  since  3  means 
are  to  be  introduced  between  1  and  9,  n  must  be  5.     Hence,  d  = 

9 1 

:=  2.     Knowing  the  common  difference,  the  terms  succeeding 

the  first  can  be  readily  formed.  They  are  3,  5,  7 ;  and  the  5  terms  of 
the  proportion  are  1,  3,  5,  7,  9. 


EXAMPLES. 

1.  Required  15  means  between  1  and  9. 

Alls,  d  =  i.     Series,  1 .  1*  .  2  .  2^  .  3  .  3^  .  4  .  4J  .  5  .  5^  .  6  .  6K 

7.7^8.81.9. 

It  is  plain  that  an  infinite  number  of  means  may  be  inserted  between 
1  and  9,  by  making  the  common  difference  indefinitely  small. 

2.  Insert  10  means  between  1  and  2. 

Ans.  d  =  j\.   Series,  1 .  l-^V  •  1 A  •  Itt  •  Itt  •  h\  ■  h\  •  Irr  •  h\  • 

1_9_.  .  llfl  .  2. 

GENERAL  EXAMPLES. 

1.  Find  4  numbers  in  arithmetical  progression,  whose  sum  is  14, 
and  the  sum  of  whose  squares  is  54.  Ans.  2,  3,  4,  5. 

Let  y  —  3a;  =  lesser  extreme,  and  let  2x  =  common  difference. 
Then  the  other  terms  are  y  —  x,  i/  +  x  and  ^  +  Sx.  Hence,  by  the 
conditions,  y  —  Bx  +  i/  —  x  -\-  i/  +  x  +  i/  +  3x^4y  =  14,  or  y  = 
3^.  And  (y  —  Sxy  +  (y  —  xf  +  (y  +  xY  +  (y  +  Sx)'  =  4/  + 
20x2  ^  54^  and  eliminating  f,  we  have  49  +  20a;'  =  54,  or  x^  =  I. 
Then,  x  =  ±  ^.     From  which  the  above  series. 

2.  Find  5  numbers  in  arithmetical  progression,  whose  sum  is  40,  and 
the  sum  of  whose  squares  is  410.  Ans.  2,  5,  8,  11,  14. 

Let  y  —  4a;  =  lesser  extreme,  and  2x  =  common  difference.  Then 
the  terms  will  be  represented  by  y  —  4a:,  y  —  2a;,  y,  y  +  2a;,  y  +  4a;. 


ARITHMETICAL    PROGRESSION.  421 

3.  Find  4  numbers  in  arithmetical  progression,  whereof  the  product 
of  the  extremes  is  700,  and  the  product  of  the  means  750. 

Ans.  20,  25,  30,  351 

4.  The  sum  of  3  numbers  in  arithmetical  progression  is  18,  and  the 
sum  of  their  squares  116.     What  are  the  numbers?      Ans.  4,  6,  8. 

Let  y  represent  the  mean,  and  x  the  common  difference,  then  will 
7/ — X,  y  andy+a;,  represent  the  three  terms  of  the  progression. 

5.  A  and  B  are  separated  by  a  distance  of  45  miles.  A  travels 
towards  B  at  the  rate  of  1  mile  the  first  hour,  2  miles  the  second  hour, 
3  miles  the  third  hour,  &c.  B  travels  towards  A  at  the  rate  of  10 
miles  the  first  hour,  8  miles  the  second  hour,  6  miles  the  third  hour, 
&c.     When  will  they  come  together  ? 

Ans.  At  the  end  of  the  fifth  hour,  and  also  at  the  end  of  the 
eighteenth  hour. 

The  second  value  needs  explanation.  The  formula  Z  =  a — {ix  —  l)c^. 
which  gives  the  distance  travelled  by  B  during  the  Ji*"  hour,  shows  that 
B  does  not  travel  at  all  the  sixth  hour.  And  since,  when  w  ^  6,  I  is 
negative,  we  conclude  that  B  turns  back,  and  travels  in  the  contrary 
direction  after  the  sixth  hour. 

15  miles 

A  P  B  M 


At  the  end  of  the  fifth  hour  the  two  travellers  were  together  at  P, 
15  miles  from  A.  In  13  more  hours  the  traveller  A  will  be  156  miles 
from  P,  at  a  point  M.  The  traveller  B  rests  during  the  sixth  hour,  and 
at  the  beginning  of  the  seventh  hour  turns  back  in  pursuit  of  A,  and 
at  the  end  of  the  eighteenth  hour  from  the  time  of  his  first  starting 
from  B,  he  will  also  be  at  M,  156  miles  from  B.  These  results  are 
easily  deduced  from  the  formulae. 

When  the  travellers  come  together  at  M,  A  will  have  travelled  17 
miles,  and  B  201  miles. 

The  problem  explains  most  satisfactorily  the  meaning  of  a  negative 
solution,  and  shows  that,  when  distance  is  the  thing  to  be  determined, 
the  negative  sign  always  implies  a  change  of  direction. 

6.  A  fugitive  from  justice  has  one  hour  the  start  of  the  officers  of 
the  law,  and  travels  uniformly  at  the  rate  of  1%  miles  per  hour.     The 
officers  travel  5  miles  the  first  hour,  6  the  second,  7  the  third,  &c.    How 
36 


422  GEOMETRICAL    PROGRESSION. 

long  will  it  be  from  tlie  time  of  tlie  fugitive's  escape  until  his  arrest 
again  ?  Ans.  9  hours,  n  =  8. 

7.  The  mean  of  seven  terms  in  arithmetical  progression  is  6,  and  the 
product  of  the  extremes  27.  What  is  the  common  diflference,  and 
what  the  series  ?  Ans.  d  =  1.     Series,  3,  4,  5,  6,  7,  8,  9. 

8.  A  liquor-seller  makes  24  gallons  of  a  mixture  of  water,  brandy, 
and  rum.  The  number  of  gallons  of  the  three  fluids  constitute  a  pro- 
gression, of  which  the  number  of  gallons  of  water  is  the  least  extreme, 
and  that  of  rum  the  greatest  extreme.  The  sum  of  the  gallons  of  water 
and  brandy  is  equal  to  the  number  of  gallons  of  rum.  Required  the 
quantity  of  each.  Ans.  4,  8,  and  12. 

9.  A  man  sold  a  horse  upon  condition  that  he  should  receive  1  cent 
for  the  first  nail  in  his  shoes,  11  cents  for  the  second  nail,  21  cents  for 
the  third  nail,  &c.,  for  the  32  nails  in  his  shoes.  How  much  did  he 
get  for  him  ?  Ans.  $49  and  92  cents. 

10.  One  hundred  stones  are  placed  on  the  same  straight  line,  4  yards 
apart.  How  far  will  a  person  have  to  walk,  who  puts  them  one  by  one 
in  a  basket  placed  on  the  same  straight  line,  4  yards  from  the  first  stone, 
and  8  yards  from  the  second  stone  :  the  person  being  supposed  to  start 
from  the  basket  ?  Ans.  22  miles,  and  1680  yards. 

11.  Same  problem  as  last,  except  that  the  basket  is  placed  at  the 
first  stone.  A7is.  22  miles,  and  880  yards. 

12.  Same  problem  as  10th,  except  that  the  carrier  is  at  the  other  end 
of  the  line,  at  the  100th  stone.  Required  the  last  term,  the  common 
difierence,  the  first  term  and  the  entire  distance  ? 

A71S.  S  =  22  miles,  1280  yards,  I  =  S,  a  =  792. 

The  progression  begins  with  the  second  term. 


GEOMETRICAL  PROGRESSION. 

437.  A  Geometrical  Progression  is  a  series  of  terms,  any  one  of 
which  is  formed  from  that  which  immediately  precedes,  by  multiplying 
by  a  constant  quantity.     When  the  constant  multiplier  is  greater  than 


GEOMETRICAL    PROGRESSION.  423 

unity,  the  series  is  an  increasing  progression.  Thus  1,  2,  4,  8,  16,  32, 
64,  &c.,  constitute  an  increasing  progression,  the  constant  multiplier  or 
ratio  of  the  progression  being  2.  And  32,  16,  8,  4,  2,  1,  is  a  decreas- 
ing progression,  with  the  ratio  \.  The  ratio  of  a  decreasing  progression 
is  always  unity  divided  by  an  entire  quantity.  It  is  evident  that  an 
increasing  progression  taken  in  reverse  order  will  be  a  decreasing  pro- 
gression, and  that  the  ratio  of  the  latter  progression  will  be  the  recipro- 
cal of  the  former.     The  converse  of  this  is  also  plainly  true.     Thus  6, 

-,  -^,  -3,  &c.,  is  a  decreasing  progression  whose  ratio  is  -,  a,   being 

supposed  an  entire  quantity,  and  -3,  -^,  -,  and  I,  constitute  an  in- 
creasing progression,  whose  ratio  is  a. 

It  is  plain  that  a  Geometrical  Progression  differs  from  a  Geometrical 
Proportion  only  in  its  number  of  terms. 


FORMULA  FOR  THE   Klh  TERM. 

438.  Let  the  first  term  be  a,  and  r  the  ratio  of  the  progression  :  then 
ar  will  be  the  second  term,  ar^  the  third,  ai^  the  fourth,  and  so  on. 
That  is,  each  term  is  equal  to  the  first  term,  multiplied  by  the  ratio 
raised  to  a  power  denoted  by  the  number  of  terms,  which  precede  the 
required  term.  Hence,  if  I  represent  the  «*■•  term,  we  must  have  I  ^ 
ar°-',  that  is,  the  n*"  term  is  equal  to  the  first  term,  multiplied  hy  the 
ratio  raised  to  the  (n — 1)  power.  I  may  represent  the  last  term,  in 
which  case  n  —  1  will  be  the  number  of  terms  less  one.     The  formula, 

I  =z  ar^^,  will  enable  us  to  determine  any  term,  when  the  number  of 
preceding  terms,  the  ratio  and  the  first  term  are  known.  When  n  =  1, 
we  have  I  =1  a  ;  when  n  ^  2,  I  =  ar,  &c.  When  a  =  0,  I  also  =0; 
wh&a.  r  =z  0,1  also  =  0  ;  when  a^r,l  =  r".  * 

EXAMPLES. 

1.  Find  the  last  term  of  the  progression,  1,  2,  4,  &c.,  carried  on  to 

II  terms  inclusive.  Ans.  I  =  1024. 

2.  Find  the  h*'^  term  of  the  progression,  3,  9,  27,  &c.       Ans.  3°. 

3.  Find  the  21st  term  of  the  progression,  2,  8,  32,  128,  &c. 

Ans.   2199023255552. 

4.  Find  the  6th  term  of  the  series,  64,  16,  4,  &c.  Ans.  Jg. 


424  GEOMETEICAL    PROGRESSION. 

The  n}^  term  being  made  up  of  a  product  and  a  power,  is  much 
greater  in  geometrical  than  in  arithmetical  progression,  when  r  is  ^  1, 
and  much  less  in  the  former  than  in  the  latter,  when  i'<^l.  Thus,  a 
corresponding  example  to  3,  in  arithmetical  progression,  with  the  first 
term  2  and  common  difference  4,  would  give  82  for  the  21st  term; 
and  the  answer  to  a  corresponding  problem  to  4,  in  arithmetical  pro- 
gression, would  be  44. 


FORMULA  FOR  THE  RATIO  OF  THE   PROGRESSION. 


439.  The  equation,  I  =  ar°~\  gives  r  =    \  / — , 


that  is,  the  ratio  is 


equal  to  the  (n  —  ly^  root  of  the  quotient  arising  from  dividing  the  to^^ 
term  hy  the  \st  term. 

This  formula  can  be  used  when  n,  I,  and  a,  are  known.     When  I  =  a, 

we  have  r  =  1 ;  when  n  =  1,  r  =  oo ;  when  n  =  2,  /•  =  — . 


EXAMPLES. 

1.  The  3d  term  of  a  geometrical  series  is  256,  and  the  first  term  4. 
What  is  the  ratio?  Ans.  r  =  8. 

2.  The  fourth  term  of  a  geometrical  series  is  2401,  and  the  first 
term  7.     Required  the  ratio.  Ans.  r  =  7. 

3.  The  first  term  of  a  geometrical  series  is  50,  and  the  fourth  term 
6250.     Required  the  ratio.  Ans.  r  =  5. 

4.  The  first  term  of  a  geometrical  series  is  6250,  and  the  fourth 
term  50.     Required  the  ratio.  Ans.  r  =  ^. 

5.  The  first  term  of  a  geometrical  series  is  a,  and  the  (m  +  1)*'' 
term  a^.     Required  the  ratio.  Ans.   y/a. 


Corollary. 

440.  The  formula,  r  =z    \  /  — ,  enables  us  to  insert  any  number  of 

means  between  two  extremes.     If  we  wish  to  introduce  m  means  be- 
tween a  and  I,  then  the  total  number  of  terms  will  be  m  -f  2.     Hence, 


GEOMETRICAL    PROGRESSION.  425 


Thus,  let  it  be 


»(  +  !  /  / 
n  —  l=m  +  2  —  \=zm  +  1,  and  r  =    Kj  — . 

required  to  insert  3  means  between  2  and  32 ;  then,  r  =  V  2*  =^  \/16 
=  2.     And  the  series  is  2,  4,  8,  16,  32. 


EXAMPLES. 

1.  Find  three  means  between  yW  and  1728,  and  the  ratio  of  the  pro- 
gression, Ans.  1  .  12  .  144,  and  r  =z  12. 


Ans.  40  and  4800. 


FORMULA  FOR  THE  SUM  OF  THE  TERMS. 

441.  If  we  multiply  a  series  of  "terms  in  Geometrical  Progression 
by  the  ratio,  a  new  series  will  be  produced,  whose  first  term  will  be  the 
same  as  the  second  of  the  old  series,  and  whose  other  terms  will  all 
be  the  same  as  the  corresponding  terms  of  the  original  series,  except 
its  last  term.  It  is  plain,  then,  that  if  the  old  series  be  subtracted 
from  the  new,  all  the  terms  will  be  cancelled  except  the  first  of  the  old 
series  and  the  last  of  the  new. 

Take,  for  example,  the  series,  3,  9,  27,  81,  243. 

And  multiply  each  term  by  the  ratio,       9,  27,  81,  243,  729. 
Subtracting  the  first  from  the  3,  729. 

second  series,  we  have  left  but  two  terms. 

Let  S  represent   the  sum   of  n  terms  in  geometrical  progression. 

Then,  S  =  a  -f  ar  +  ar^  +  ar^ or""'. 

Multiply  by  r,  and  we  have  Sr  =  ar  +  ar'^  +  ar^ ar^~^  -f  ar^. 

Hence,  Sr  —  S  =z  ar^  —  a,  and  S  =  — ^^ = — . 

r  —  1 

That  is,  the  sum  of  the  terms  is  equal  to  the  first  term  into  the  diffe- 
rence hctioeen  unity  and  the  ratio  raised  to  a  power  denoted  hy  the 
number  of  terms,  and  this  product  divided  hy  the  ratio,  less  1.  This 
formula  can  be  used  when  the  first  term,  the  ratio,  and  number  of 
terms   are   known,   or   can   be    determined.       Since  I  =  ar'''^  then, 

Ir  z=  ar^,  and  the  formula  may  be  written  S  =z ^  ;    that  is,  the 


426  GEOMETRICAL    PROGRESSION. 

sum  of  the  terms  is  equal  to  the  last,  or  n"",  term  into  the  ratio,  minus 
the  first  term,  and  this  difference  divided  hij  the  ratio,  less  1. 

When  n  =  0,  the  second  members  of  the  equations  in  S  and  Sr  will 
cancel  each  other,  and  give  S  =  Oj  but  when  r  =  1,  the  equation  in 

S  becomes  S  =  na,  whilst  the  formula,  S  =  — ^ ~,  becomes  S  =  g. 

This  is  generally  the  symbol  of  indetermination,  but,  in  the  present  in- 
stance, indicates  a  vanishing  fraction.  The  common  factor  to  the  two 
terms  of  the  fraction  is  plainly  r  —  1,  because  r  —  1  is  an  exact  divisor 

of  the  numerator.     Performing  the  division,  we  have  S  =  — — "~    ^ 

=  a  (r""'  +  r"-^  +  r°~^  + r  +  1).     Now,  making  r  =  l, 

we  have  S  =  7ia 

442.  When  the  progression  is  decreasing,  8r  is  less  than  S ;  hence, 
to  find  S,  we  must  take  Sr  from  S.     Then,  S  —  Sr  =  a  —  ar",  or 

ft    /I    I  7'^'^  ft       1_  I  ly* 

S  =  — !q =  — = .  that  is,  the  sum  of  the  terms  of  a  de- 

1  —  r  1  —  r  •' 

creasing  progression  is  equal  to  the  first  term  into  unity,  minus  the  last, 

or  n"",  ter7n  into  the  ratio,  and  this  difference,  divided  hy  one,  minus 

the  ratio. 

EXAMPLES. 

1.  The  first  term  is  2,  the  ratio  2,  and  the  number  of  terms  9. 
What  is  the  sum  of  the  terms  ?  Ans.  1022. 

2.  The  first  term  is  512,  the  last  term  2.     Required  the  sum  of  the 
9  first  terms  of  the  progression.     »  Ans.  1022. 

3.  Required  the  sum  of  the  series,  1  +  x  -\- x^  -^  x^ cc"-'. 

a;"  — 1 


x  —  1 

4.  Required  the  sum  of  the  series,  a;""'  +  cc""^  +  ....  a;  +  1. 

Ans. 


1  —  x 

5.  Required  the  sum  of  the  series,  cc""'  +  a;"~^y  -{-  x'"~^y^  .  .  . 

^  ^  A71S.  ^ ,  or ^ 

y—x  x—y 

6.  Required  the  sum  of  the  series,  x"^  -f  ax  +  a^. 

.         a^  —  x^        x^  —  rr 
Ans.   ,  or 


GEOMETRICAL    PROGRESSION.  427 

("r"  —  1) 

443.  The  two  equations,  I  =  ar''~\  and  s  =  a =— ,    contain 

five  quantities,  any  two  of  which  can  be  determined  when  the  other 
three  are  known.     Thus,  when  a,  I  and  n,  are  known,  we  obtain,  by 

combining  these  equations,  and  eliminating  r,  — 


ive  get  a  =  -^ 


In  like  manner,  by  eliminating  ?,  we  get  a  =  -   ^ ■ ;  and,  by  eli- 
minating a,  we  find  I  = 


AN  INFINITE  DECREASING  PROGRESSION. 

444.  The  formula  for  the  sum  of  the  terms  of  a  decreasing  progres- 
sion takes  a  remarkable  form  when  the  series  is  infinite. 

For,  S  =  t; may  be  written,  S  =  -z . 

1  —  r        ''  1  —  r       1  —  r 

Now,  since  r  is  a  fraction  whose  numerator  is  unity,  like  the  fraction 

r°         . 
1,  it  is  plain  that,  when  «.  =  oo  ,  r°  will  be  zero.     Hence,  = will 

be  zero,  and  that  part  of  the  formula  may  be  neglected  when  the  num- 
ber of  the  terms  is  infinite,  and  we  may  write  S  =  z: ;  that  is,  the 

1  —  r 

sum  of  the  terms  of  an  infinite  decreasing  progression  is  equal  to  the 
first  term  divided  bi/  unity,  minus  the  ratio  of  progressions. 

This  formula  will  give  the  limit  of  the  series,  that  is,  a  value  which 
the  sum  of  the  terms  cannot  exceed.  Thus,  the  decimal  -33333,  &c., 
may  approach  infinitely  near  to  -i,  but  can  never  exceed  it. 


EXAMPLES. 

1.  Find  the  limit  of  the  value  of  the  decimal,  -666666,  &c. 


For  a  =  -j%,  and  r  =  ^\;  hence,  S  =       '"  ,    =  «  =  |. 

■••       To 

2.  Find  the  limit  of  the  value  of  the  decimal,  •1111111,  &c. 

Ans.  ^. 

3.  Find  the  limit  of  the  value  of  the  decimal,  -55555,  &c. 


428  GEOMETRICAL    PROGRESSION. 

4.  Find  the  limit  of  the  value  of  -99999,  &c.  Ans.  1. 

5.  Find  the  limit  of  the  value  of  -88888,  &c.  A71S.  |. 

6.  Find  the  limit  of  the  value  of  the  series,  1  +  |  +  t  s  +  bV  +  ^^^ 

Ans.  l+i 

7.  Find  the  limit  of  the  value  of  J  +  j\  +  g^j  &c.         Ans.  f. 

8.  Find  the  limit  of  the  value  of  the  series,  20  +  4  +  |  +  5%  +  &c. 

Ans.  25. 

9.  Find  the  limit  of  the  value  of  f  +  -^^  +  ji-g  +  &c. 

A71S.  1. 

10.  How  much  will  the  sum  of  the  series,  |  +  ^-^,  &c.,  differ  from 
unity,  when  four  terms  only  are  taken  ?  Ans.  By  g^^ 

Use  the  expression,  = ,  which  can  only  be  neglected  when  w  =  oo  . 

11.  Find  the  sum  of  the  series,  42  +  6  +  f  +  4^^+  -gfj,  &c. 

^  Ans.  49. 

12.  Find  the  limit  of  the  value  of  6  -f  f  +  /^  +  ^fg  +  &c. 

Ans.  7. 

13.  By  how  much  will  the  sum  of  the  series,  ^  +  ^  -h  4%  +  3I3  -f- 
&c.,  differ  from  7,  when  three  terms  only  are  taken. 

Ans.  By  ^\. 

GENERAL   EXAMPLES. 

1.  The  first  term  of  a  geometric  progression  is  4,  the  last  term  32, 
and  the  number  of  terms  4.     Required  the  sum  of  the  terms. 

Ans.  60. 

2.  There  are  three  quantities  in  geometric  progression,  their  sum  is 

a,  and  the  sum  of  their  squares,  b^.     What  are  the  quantities  ? 

a^  _  J2  ^2  +  52       ^lOa'b^  —  Sb'  —  Sa* 

Ans.  y  =  — —— — ,  X  ^  — -. -. ,  z  = 

"^  2a     '  4a      ^  4a  ' 


^lOa^P  —  Sb'  —  Sa' 


4a  4a 

For,  let  X,  y  and  z,  denote  the  three  quantities.  Then,  x  -^y  -\-  z 
z=a,y'^=xz,  and  x^  -\- y^  +  z^  ■=  W.  Transposing  y  to  the  second 
member,  and  squaring  the  first  equation,  we  get  x^  +  2xz  +  z^^a^  — 
lay-^-y^^  which,  combined  with  the  second,  gives  cc^  +  ^  +  2^  =  a^  — 


GEOMETRICAL    PROGRESSION.  4^ 

2ay.     Subtract  the  third  equation  from  this,  and  solve  with  reference 

0,2 52 

to  y;  we  will  find  then,y  =  — ^ .     The  third  equation  gives  x^  ■\-  ^ 

=  h^  — y^.     Subtract  from  this  the  equation,  2xz  =  Ix^.     Then,  01?  — 

2a;2;  ■\-  ^  =.W-  —  3^^,  and,  taking  the  square  root  of  both  members,  we 

find,  X  —  z  ■=■  s/i>^  —  3y  (M) ;  but,  from  the  first  equation,  x  -\-  z  =■ 

a  —  y  (N).     Adding  and  substracting  (M)  from  (N),  we  get  x  = 

a—y^  ^h^  —  ^f       a?  +  1/       ^l^a'U"  —  Sh*  —  S^' 

=  — ] 2 ,  and  z  — 

2  4a  4a 


a—y  —  ^b'  —  Sy^   _  ^'  +  ^' _  -v/ lt'«'^'  —  3i'  —  3a* 
2  ^      4a  ~  I^  ■ 

The  problem  might  have  been  solved  by  the  ordinary  method  of  eli- 
mination, but  it  would  have  led  to  an  equation  of  the  fourth  degree. 

3.  Two  travellers  were  separated  by  a  distance  of  36  miles.  The 
one  in  advance  travelled  1  mile  the  first  day,  5  miles  the  second  day, 
9  the  third  day,  and  so  on.  The  one  in  the  rear  travelled  1  mile  the 
first  day,  and  increasing  his  rate  in  a  geometrical  ratio,  travelled  64 
miles  on  the  seventh  day,  when  he  overtook  the  advanced  traveller. 
Required  the  uniform  rate  of  increase  of  the  second  traveller's  daily 
distance.  Ans.  r  =  2  miles. 

4.  Find  r  when  s,  a  and  I  are  known. 


s—r 


Problem  3  is  but  a  particular  case  of  problem  4. 

5.  There  are  four  quantities  in  geometric  progression,  the  sum  of  the 
first  two  is  a,  and  of  the  last  two,  b.  What  is  the  ratio  and  what  are 
the  numbers  ?  

.                    /X     T^T      ,                  a                 V   a                h 
Ans.  r  =\  /  — .     Numbers,  =1,  =:, =, 

and 


6.  There  are  four  numbers  in  geometric  progression,  the  sum  of  the 
first  two  is  30,  and  of  the  last  two,  750.  What  is  the  ratio,  and  what 
are  the  numbers  ?         Ans.  r  =  5.     Numbers,  5,  25,  125  and  625. 


430  GEOMETRICAL    PROGRESSION. 

7.  There  are  three  quantities  in  geometric  progression,  the  first  is  a, 
and  the  second  h.     Required  the  third.  .         h^ 

a 

8.  There  are  three  numbers  in  geometric  progression,  the  first  is  4, 
and  the  second  20.     Required  the  third.  Ans.  100. 

9.  Find  a,  when  ?,  r  and  S  are  known. 

Ans.  a  =  S  —  (S  —  T)r. 

10.  The  ratio  of  a  geometric  progression  is  5,  the  sum  of  the  terms 
780,  and  the  last  term  625.     Required  the  first  term.  Ans.  5. 

11.  Find  I,  when  a,  r  and  S  are  known. 

Ans.  I  =  S . 


12.  A  rich,  but  charitable  man,  gave  $20  to  the  American  Tract 
Society,  twice  as  much  to  the  Board  of  Foreign  Missions,  four  times  as 
much  to  the  Board  of  Domestic  Missions,  and  so  on  in  the  same  ratio. 
His  last  contribution  was  to  his  own  church,  and  his  entire  charity 
amounted  to  $1020.     How  much  did  he  give  his  church  ? 

V  Ans.  $520. 

13.  A  man  sold  a  horse  upon  the  following  terms  :  He  was  to  receive 
one  cent  for  the  first  nail  in  the  horse's  shoes,  two  cents  for  the  second 
nail,  four  cents  for  the  third,  and  so  on  for  the  32  nails  in  the  horse's 
shoes.     Required  the  price  of  the  horse. 

Ans.  4,294,967,295  cents. 

14..  A  gentleman  made  a  donation  to  a  charitable  institution,  upon 
the  following  conditions  :  Five  hundred  dollars  were  to  be  received  the 
first  year,  $250  the  second  year,  $125  the  third  year,  and  so  on  forever. 
Required  the  amount  of  the  donation.  Ans.  $1000. 

15.  A  gentleman  wishes  to  bequeath  $3000  to  a  Benevolent  Associa- 
tion in  the  following  manner  :  Two  thousand  dollars  to  be  given  the 
first  year  to  meet  the  present  wants  of  the  association,  and  the  donation 
every  year  after,  forever,  to  be  in  a  decreasing  ratio.  What  must  the 
ratio  be  ?  Ans.  J. 

16.  A  gentleman  invested  $1000  for  the  benefit  of  a  charitable 
institution,  so  that  one  half  less  should  be  drawn  from  it  each  year  than 
the  preceding  year,  forever.     How  much  must  be  drawn  the  first  year  ? 

Ans.  $500. 


INEQUALITIES.  431 


INEQUALITIES. 

445.  An  inequality  is  an  expression  to  signify  that  two  quantities  are 
unequal  to  each  other.  Thus,  A  ^  B,  is  an  inequality  indicating  that 
A  is  greater  than  B.  For  equalities,  it  matters  not  on  which  side  of  the 
sign  of  equality  the  two  quantities  are  written.  Thus,  A  =  B  may  be 
written  B  =  A.  But,  for  inequalities,  there  can  be  plainly  no  trans- 
position of  the  two  quantities  without  a  corresponding  inversion  of  the 
sign.  In  the  expression,  A  ^  B,  if  A  be  changed  to  the  second  mem- 
ber, and  B  to  the  first,  there  must  be  an  inversion  of  the  sign.  For,  let 
m  be  the  quantity  which,  added  to  B,  makes  it  equal  to  A,  then,  A  = 
B  +  m,  or  B  +  m  =  A.  Then  B  <^  A.  So,  B  and  A  have  changed 
places,  with  a  corresponding  inversion  of  sign. 

Inequalities  are  used  to  determine  the  limits  between  which  quantities 
are  found,  and  are  of  frequent  application  in  the  higher  mathematics. 
As  a  simple  illustration,  suppose  we  have  the  two  inequalities,  a;  ^  5, 
and  X  <^  10.  We  see  that  x  must  be  some  number  between  5  and  10, 
and  these  numbers  are  the  Umifs  to  its  values.  When  the  enunciation 
of  the  problem  restricts  the  solution  to  positive  quantities,  there  are  but 
two  limits,  the  superior  positive  limit,  and  the  inferior  positive  limit. 
So,  if  the  solution  be  restricted  to  negative  quantities,  there  are  but  two 
limits,  the  superior  negative  limit,  and  the  inferior  negative  limit.  lYi 
general,  however,  there  are  four  limits,  two  superior  and  two  inferior. 

Two  inequalities  are  said  to  subsist  in  the  same  sense,  when  the 
greater  quantity  lies  on  the  same  side  of  the  sign  in  both  of  them;  and 
they  subsist  in  a  contrary  sense,  when  the  greater  quantity  in  one  ine- 
quality lies  on  a  different  side  of  the  sign  from  that  occupied  by  the 
greater  quantity  in  the  other  inequality. 

Thus,  4^2,  and  5^3,  subsist  in  the  same  sense;  so  also,  2  <^  4, 
and  3  <^  5.     But  4^2,  and  3  <^  5,  subsist  in  a  contrary  sense. 

When  two  quantities  are  negative,  that  one  is  the  least,  algebraically, 
which  contains  the  greatest  number  of  units.     Thus,  —  15  <C  —  l^j  ' 
-5>-7. 

446.  There  are  ten  important  principles,  belonging  to  the  subject  of 
inequalities,  which  require  to  be  demonstrated. 

1.  If  the  same  quantity  be  added  to,  or  subtracted  from,  the  two  mem- 
bers of  an  inequality,  the  resulting  inequality  will  subsist  in  the  same 
sense. 


432  INEQUALITIES. 

For,  let  A  ^  B,  and  m  the  difference  between  A  and  B,  then,  A  = 
B  +  m.  Now,  add  or  subtract  the  same  quantity,  c,  from  both  members, 
and  there  results  Adbc  =  B  +  m=bc.  Hence,  of  course,  A  ±  c  ^ 
B  ±  c,  and  the  inequality  subsists  in  the  same  sense. 

This  principle  enables  us  to  transpose  terms  from  one  member  of  the 
inequality  to  the  other.  Take,  x  -\-  2  ^  a  ;  add  —  2  to  both  members, 
and  there  results  x^  a  —  2.  So,  transposition  is  effected  in  inequalities, 
as  in  equations,  by  a  change  of  sign. 

2.  If  the  two  members  of  an  inequality  are  multiplied  by  a  positive 
quantity,  the  resulting  inequality  will  subsist  in  the  same  sense. 

For,  let  A  ^  B,  and  A  =  B  +  m.  Multiply  both  members  by  c, 
and  we  get  Ac  =  Be  -f  mc.  Hence,  Ac  ^  Be.  This  principle  serves 
to  free  the  fractions,  if  any,  of  their  denominators. 

Take  -^ -r-'^l/.     Clearing  of  fractions,  there  results  2x  —  a  ^ 

A1j\ 

3.  If  both  members  of  an  inequality  are  divided  by  a  positive  quan- 
tity, the  resulting  inequality  will  subsist  in  the  same  sense. 

For,  let  A  ]>  B,  and  m  the  difference  between  A  and  B.    Then  A  = 

B  +  m.     Dividing  both  members  by  c,  there  results,  —  = 1 . 

Hence,  — ■>■ — . 

The  last  principle  serves  to  clear  the  unknown  quantity  of  its  coeffi- 
cient when  positive.     Take,  2a;  ^  a  +  4&^,  then,  x  ^  — . 

Li 

The  three  foregoing  principles  serve  to  solve  inequalities,  when  the 
coefficient  of  the  unknown  quantity  can  be  made  positive. 

4.  If  two  inequalities,  subsisting  in  the  same  sense,  are  added 
together,  member  by  member,  the  resulting  inequality  will  subsist  in 
the  same  sense. 

For,  let  A  >  B,  and  C  >  D.  Let  m  be  the  difference  between  A 
and  B,  and  n  the  difference  between  C  and  D.  Then,  A  =  B  +  m, 
and  C  =  I)  -f  n.  Adding  the  two  equations,  member  by  member, 
there  results  A  +  C  =  B  +  D  +  m  +  «.     Hence,  A  +  C  >  B  +  D. 

5.  If  an  inequality  be  multiplied  by  a  negative  quantity,  the  result- 
ing inequality  will  subsist  in  a  contrary  sense. 

For,  let  A  ^  B,  or  A  =  B  +  m.  Multiply  both  members  by  —  c, 
and  there  results  —  Ac  =  —  Be  —  w,  or  m  —  Ac  =  —  Be.     Hence, 


INEQU..LITIES.  433 

Ac  being  numerically  greater  than  Be,  we  have,  algebraically,  —  Ac  <^ 
—  Be. 

The  foregoing  principles  enab'e  us  to  eliminate  between  inequalities 
as  between  equalities,  when  we  have  two  inequalities  involving  two  un- 
known quantities  whose  signs  are  known,  and  that  of  one  of  them  in 
one  inequality  contrary  to  that  of  the  same  unknown  quantity  in  the 
other  inequality. 

Take  x  +  y  >  12, 

Adding  member  by  member  we  get  x^  10,  and  this  substituted  in 
the  first  equation,  gives  y  >  2.  Let  x  =  10  +  m.  Then,  from  the 
2d  equation  i/  <^2  +  m.  The  value  of  i/  is  then  fixed  between  the 
limits  of  2  and  2  +  m. 

6.  If  two  inequalities,  subsisting  in  the  same  sen.-:  •,  are  subtracted 
member  by  member,  the  resulting  inequality  may  subsist  in  the  same  or 
a  contrary  sense. 

For,  let  A  >  B,  and  C  >  D.  Then,  A  =  B  +  w,  and  C  =  D  +  ». 
Subtracting  member  by  member,  there  results  A  —  C  =  B  +  m  — 
(D  +  n);  or.  A  — C  =  B  — D  +  m  — «. 

Now,  if  m  ^  «,  A  —  C  will  be  equal  to  B  —  D  increased  by  a  posi- 
tive quantity,  and,  of  course,  will  be  greater  than  it.  But,  if  71  "^  m, 
A  —  C  will  not  be  equal  to  B  —  D,  until  it  has  been  diminished  by  the 
diflference  between  n  and  m. 

In  that  case,  A  —  C  <  B  —  D. 

Take,  8  >  4,        m  =  4, 

6  >  3,         n  =  3. 
8— 6>1. 

The  resulting  equality  subsists  in  the  same  sense,  since  m  >  n. 

But,  take  8  >  6,        m  =  2, 

6  >  1,  n  =  5. 

8  —  6  <  5. 

The  resulting  inequality  subsists  in  a  contrary  sense,  since  n  >  m. 

8.  If  both  members  of  an  inequality  are  essentially  positive,  they 
may  be  squared  without  altering  the  sense  of  the  inequality. 

For,  let  A  >  B,  or  A  =  B  -f  m,  then  A^  =  B^  +  2Bm  +  m^. 
Hence,  A^  >  B^ 

37  2c 


434  INEQUALITIES. 

9.  If  the  two  members  of  an  inequality  have  contrary  signs,  the  re- 
sulting inequality,  after  squaring,  may  silhsist  in  the  same  or  a  contrary 


For,  let  A  >  —  B,  or  A  =  —  B  +  m.  Then,  A'  =  B^  —  2Bm  + 
m^  If  m^  >  2Bm,  then,  A^^B^;  for,  A''  is  equal  to  B^,  increased 
by  the  difference  between  m^  and  2Bm.  But,  if  ni^  <^  2B?k,  then,  A^ 
<^  B'^  by  the  difference  between  2Bm  and  m^. 

Take,       3  >  —  2 ;  then,  m  =5,  m^  =  25,  and  2Bm  =  20. 
Squaring,  we  have,  9  >  4,  since  m^  >  2Bm. 
But  take,  3  >  —  6 ;  then  m  =  9,  m^  =  81,  and  2Bm  =  108. 
Squaring,  we  get,  9  <C  36,  an  inverted  sign,  since  m^  <  2Bm. 

When,  therefore,  the  sign  of  either  member  is  unknown,  it  is  not 
permitted  to  square  the  inequality. 

10.  When  the  two  members  of  an  inequality  are  divided  by  a  nega- 
tive quantity,  the  resulting  inequality  will  subsist  in  a  contrary  sense. 

For,  let  A  ^  B,  or  A  =  B  +  m.     Dividing  both  members  by  —  c, 

there  results = .     Since is  numerically  greater 

c  c         c  c 

than ,  it  must  be  algebraically  less. 

c 

For  equations  of  high  degrees  there  are  generally  four  limits :  a  supe- 
rior positive  and  a  superior  negative  limit,  and  an  inferior  positive  and 
an  inferior  negative  limit.  But  the  conditions  of  the  problem  may  re- 
strict the  solutions  to  two  limits,  and  even  to  one  limit. 

A  single  equation  of  the  first  degree  can  have  but  one  limit,  this  may 
be  superior  or  inferior,  positive  or  negative.  Limits  are  determined  by 
means  of  inequalities. 


EXAMPLES. 

1.  The  value  of  x  in  an  equation  is  such  that,  twice  the  value  in- 
creased by  unity  cannot  be  less  than  7.  What  is  the  lower  limit  of 
the  value  ?  We  have,  from  the  conditions,  2a;  -f  1  ^  7.  Hence,  a;  ^  3. 
The  inferior  positive  limit  is  then  3,  and  any  number  above  3  may  be 
the  value  required. 

2.  The  value  of  cc  in  a  simple  equation  of  the  first  degree  is  known 
to  be  such,  that  twice  the  value,  plus  unity,  is  greater  than  7,  and  that 


INEQUALITIES.  435 

three  times  the  value,  diminished  by  4,  is  less  than  11.     Between  what 
limits  does  the  value  lie  ? 

Ans.  Superior  positive  limit,  5;  inferior  positive  limit,  3. 

If  the  value  be  known  to  be  a  whole  number,  4  must  be  that  number. 

3.  A  person  desirous  of  giving  some  cents  to  a  certain  number  of 
beggars,  found  on  examination  that,  to  give  them  three  cents  apiece, 
would  require  more  than  double  of  what  he  had  about  him,  and  that,  to 
give  them  2  cents  apiece,  would  require  more  than  the  difference  be- 
tween all  he  had  and  35  cents.  How  many  beggars  were  there,  and 
how  many  cents  had  the  person  ? 

From  the  conditions,  3a;  ^  2y, 
and,  2a;  ^35 — y. 

Multiplying  the  second  inequality  by  2,  and  adding  member  by  men 
ber,  there  results,  7x  ^  70,  or  a;  ^  10.  This  value  substituted  in 
the  first  inequality,  gives  y  <C  15  +  I";  ^  representing  the  excess  of 
X  over  10.  The  value  of  x,  substituted  in  the  second  inequality,  gives 
y  >  35  —  20  —  2m;  or  y  >  15  —  2m.  Suppose  m  =  J,  then. 
y  <  151,  and  7/  >  14. 

4.  A  person  desiring  to  give  some  money  to  11  beggars,  found  that, 
to  give  them  3  cents  apiece,  would  require  more  than  twice  as  much 
money  as  he  had  about  him,  and  that,  to  give  them  2  cents  apiece, 
would  require  more  than  the  difference  between  37  cents  and  the  num- 
ber of  cents  about  him.     Required  the  number  of  cents  he  had. 

A71S.  X  <[  16  J,  and  a;  ^  15;  hence,  x  =  16. 

The  nature  of  the  problem  makes  the  solution  exact  in  this  case,  but 
it  is  very  seldom  so. 

5.  Find  the  negative  limits  of  x  in  the  inequalities, 

cc-f  3<  — 5, 
a;  —  3  >  —  7. 

Ans.  x<^  —  8,  and  x^  —  4. 

6.  Find  the  limit  of  the  value  of  x  in  the  inequality,  ax+b  -^  c 

-yd^fx.  d—l  —  c 

Ans.  Inienor  limit,     r— 

«+/ 


436         GENERAL  THEORY  OF  E(^UATIONS, 


GENEKAL  THEORY  OF  EQUATIONS. 

447.  The  general  theory  of  equations  has  for  its  object,  the  investi- 
gation of  properties  common  to  equations  of  every  degree  and  of  every 
form. 

We  will  confine  ourselves  mainly  to  the  examination  of  equations  in- 
volving but  one  unknown  quantity. 

Tke  most  general  form  of  an   equation  of  the  m*''  degree  with  one 

unknown  quantity,  is  »"•  -f  Px-"-'  +  Qx'"-^  +  Rx"-'' -f  Tie 

-f  U  =  0. 

The  coefficient  of  the  first  term  is  plus  unity,  and  the  other  coeffi- 
cients, positive  or  negative,  entire  or  fractional,  rational  or  irrational. 

A  value  has  been  defined  to  be  that  which,  substituted  for  the  un- 
known quantity,  will  make  the  two  members  equal  to  each  other.  Since, 
in  the  general  equation  of  the  m*''  degree,  the  second  member  is  zero,  a 
value  substituted  for  x,  must  reduce  the  first  member  to  zero  also. 
Hence,  in  our  discussion,  we  may  define  a  value  to  be  that  which,  sub- 
stituted for  the  unknown  quantity  in  the  equation,  will  reduce  the  first 
member  to  zero. 

GENEKAL  PROPERTIES  OF  EQUATIONS. 
First  Property. 

448.  If  any  quantity,  a,  be  a  value  of  x  in  the  equation,  a;"  +  Pa;""' 
+  .  .  .  Taj  -f  U  =  0,  the  first  member  of  this  equation  will  be  exactly 
divisible  by  a;  —  a. 

Fffr,  let  Q'  be  the  quotient  resulting  from  the  division  of  the  first 
member  by  a:;  —  a,  and  let  R'  be  the  remainder,  if  any,  after  division. 
We  shall  then  have  the  identical  equation,  x""  -f  Pa;""'  -f  Qx'"~^+  .... 
-f-  Ta;  -f  U  =  Q'  (a;  —  a)  -f  R'.  But,  for  x  =  a,  the  first  member  is, 
by  hypothesis,  equal  to  zero,  and  the  second  member  reduces  to  R'. 
Hence,  we  have  0  =  0  +  R',  and,  therefore,  R'  =  0,  and  the  division 
is  exact. 

Second  Property. 

449.  If  the  first  member  of  an  equation,  of  the  form  of  a:"  +  Px"^' 
+  Qx"-^  -t-  .  .  .  .  Tx  -f-  U  =  0,  be  divisible  by  x  —  a,  then  a  will  be 
a  value  in  this  e(juation. 


GENERAL  THEORY  OF  EQUATIONS.         437 

For,  calling  Q'  the  quotient,  we  will  have  x^  +  Px""'  -j-  Qx""^  -f- 
.  .  .  .  Ta;  +  U  =  Q(x  —  a).  Hence, then, Q'(.r  —  a)  =  0.  But,  when 
the  product  of  two  factors  is  equal  to  zero,  the  equation  can  he  satisfied 
by  placing  either  factor  equal  to  zero.  We  have  then  a  right  to  place 
X  —  a  =  0,  from  which  there  results  x  =  a.  Therefore,  a  is  a  value 
in  the  equation,  Q'(aj  —  «)  =  0,  and,  consequently,  in  the  given 
equation. 

Corollary. 

1st.  It  follows  that,  in  order  to  ascertain  whether  any  polynomial  is 
divisible  by  x  —  a,  we  have  only  to  substitute  a  for  x,  wherever  x 
occurs,  and  see  whether  the  polynomial  reduces  to  zero. 

2d.  Hence,  also,  if  a  polynomial  is  divisible  by  x  —  a,  a  will  be  a 
value  in  the  equation  formed  by  placing  the  polynomial  equal  to  zero. 

3d.  We  may  also  diminish  the  degree  of  an  equation,  when  we  know 
one  or  more  of  its  values,  by  dividing  its  first  member  successively  by 
the  binomial  factors  corresponding  to  these  values.  The  division  by 
each  binomial  factor  will  reduce  the  degree  of  the  equation  by  unity. 

4th.  Numerical  equations  can  frequently  be  solved  by  means  of  the 
first  two  properties.  Literal  equations  can  also  be  solved  in  the  same 
way,  but  more  rarely :  we  have  only  to  ascertain  what  number  or  quan- 
tity will  satisfy  the  equation ;  then,  by  dividing  out  the  binomial  factor 
corresponding  to  this  value,  we  will  have  a  now  equation  of  a  degree 
lower  by  unity.  A  second  value  may  be  found  frdm  this  equation,  and 
the  factor  corresponding  to  it  divided  out.  Thus,  we  may  continue  tho 
process  until  the  given  equation  is  reduced  to  one  of  the  second  degree, 
which  can  be  solved  by  known  rules. 

EXAMPLES. 

1.  Solve  the  equation,  x^  —  Gx-^  +  11^  —  6  =  0. 

We  see  that  a:  =  1  will  satisfy  this  equation.  Hence,  x  —  1  is  a 
divisor.  Dividing  by  x  —  1,  we  get  a  quotient,  x^  —  5x  -f  6  =  0,  by 
solving  which  we  get  a;  =  2  and  3.  Hence,  the  three  values  are  1, 
2,  and  3. 

2.  Solve  the  equation,  x*  —  bx^  -f  4  =  0. 

We  find  +1  to  be  a  value,  and,  dividing  by  a;  —  1,  we  get  a  new 
equation,  x^  -f  x^  —  4x  —  4  =,    which   is    satisfied    for   x  =  —  1. 
Dividing  by  a;  +  1,  we  get  a  new  equation,  x^  —  4  =  0,  which  gives 
37* 


438  GENERAL    THEORY    OF    EQUATIONS. 

the  two  values,  x=  -\-  2,  and  —  2.     The  four  values  are  then,  +  1, 
—  1,  H-  2,  and  —  2. 

3.  Solve  the  equation,  x^  +  Sx*  —  5x'  +  4cc  +  12  —  ISa;^  =  0. 

Ans.   +  1,  —  1,  +  2,  —  2,  and  —  3. 

4.  Solve  the  equation,  x*  —  a^x^  —  l^x^  +  a^h'^  =  0. 

A71S.    +  a,  —  a,  -f  5,  —  h. 

5.  Solve  the  equation,  of  —  a^x'^  —  I'^x^  +  a^x^  +  I'^x^  —  x'^  +  a^¥x  — 
aW  =  0.  Ans.   +  a,  —  a,  +  b,  — b,  and  +  1. 

6.  Solve  the  equation,  sc^  +  x^  —  4x  —  4  =  0. 

Ans.   +  2,  —2,  —1. 


Third  Property. 

449.  Every  equation,  with  one  unknown  quantity,  has  as  many 
values  for  this  unknown  quantity  as  is  denoted  by  the  degree  of  the 
equation,  and  has  no  more. 

Let  us  assume  the  equation,  a:;"  +  Pa;"""'  +  Qx""^  +  .  .  .  Ta;  +  U 
=  0,  which  is  of  the  m*''  degree.  It  is  to  be  shown  that  it  has  m 
values  and  no  more.  We  will  also  assume  that  every  equation  has  at 
least  one  value.     Let  a  be  one  value,  then  x  —  a  is  a  divisor.     Let 

jpm-i  _j_  p'-^m-z  _f.  Q'a;"-'  + T'ic  +  U',  represent  the  quotient  of 

the  division  of  the  first  member  by  cc  —  a.  Then  the  given  equation 
will  assume  the  form,  {x  —  a)  (x"-'  +  P'x"""^  -f  Q'x"""^  +  &c.)  = 
0  .  (A),  in  which,  the  coefficients,  P',  Q',  are  different  from  the 
original  coefficients,  P,  Q,  &c.  Now,  since,  in  equation  (A),  we  have  the 
product  of  two  factors  equal  to  zero,  we  have  a  right  to  place  either 
equal  to  zero,  let  us  place  a;-""'  +  Fx"-^'  +  Q'cc"-^  +  &c.  =  0  (B). 
Equation  (B)  will  also  have  one  value ;  suppose  it  negative  and  equal  to 
—  h.  Then,  by  the  first  property,  x  —  (  —  Z))  =  x  ^-  &  will  be  a 
divisor.  The  first  member  of  (B)  can  then  be  decomposled  into  two 
factors,  one  of  which  will  be  a;  +  &,  and  the  other  the  quotient  arising 
from  the  division  of  the  first  member  by  a;  +  &•  Hence,  (B)  becomes 
(a;  +  h)  (a;'"-^  +  F'a;'"-^  +  g'tt""-"  +  &c.)  =  0  (C).  And  (A)  can  be 
put  under  the  form  of  {x  —  a)  (x  +  I)  (x'"-^  +  T"x^-^  +  Q"x"'-^  + 
&c)  =0.  It  is  plain  that,  by  putting  the  second  factor  of  (C)  equal  to 
zero,  we  will  get  another  value,  and,  consequently,  another  divisor.    By 


GENERAL    THEORY    OF    EQUATIONS.  439 

continuing  this  process,  the  degree  of  a;  in  the  successive  quotients  will 
be  diminished  by  unity  each  time,  and,  after  m  —  1  divisions,  we  will 
obtain  a  quotient  of  the  first  degree  in  x,  from  which,  of  course,  one 
value  of  X  will  be  found.  "We  supposed  the  first  value  positive,  and 
the  second  negative,  let  us  attribute  the  double  sign,  ± ,  to  the  remain- 
ing values,  c,  d,  e,  &c.,  since  they  may  be  either  positive  or  negative. 
The  first  member  of  the  given  equation  can  then  be  decomposed  into  m 
binomial  factors  of  the  first  degree  in  x,  and  we  will  have,  a;"  +  Px"-' 

+  Qx'"-'  + +Tx  +  V=  (X  — a)  (x  +  b)  (xzpc-)  (x^zd) 

&c.  =  0. 

But,  since  each  factor  corresponds  to  a  value,  and  since  there  are  m 
divisors,  or  factors,  there  must  be  ?n.  values. 


Scholium. 

1st.  Had  we  known  that  the  first  member  of  an  equation  of  the  wi*"" 
degree  could  be  decomposed  into  m  binomial  factors  of  the  first  degree 
with  respect  to  x,  we  could  readily  have  shown  that  the  equation  must 
contain  ??i  values.  For,  it  can  be  satisfied  in  m  ways,  by  placing  each 
of  its  7)1  factors  equal  to  zero,  and  each  factor,  so  placed,  will  give  a 
value.  Moreover,  since  the  equation  can  be  satisfied  only  in  m  ways, 
it  contains  no  more  than  m  values. 

2d,  It  does  not  follow  that  all  the  values  must  be  difi"erent.  Any 
number  of  them,  even  all  of  them,  may  be  equal.  Thus,  the  equation, 
a?  —  2a;  +  1  =  0,  contains  two  values,  each  equal  to  +  1.  The  equa- 
tion, (x  —  a)""  =  0,  contains  m  values,  each  equal  to  -j-  a  The  equa- 
tion, (x  —  a)'°(x-{-by=0,  contains  m  values,  each  equal  to  +  a,  and  n 
values,  each  equal  to  —  b.     And  so  for  other  equations. 

3d.  An  equation  of  the  third  degree,  such  as  (x  —  a)  (x  —  b)  (x  —  c) 
=  0,  will  contain  three  divisors  of  the  first  degree,  {x  —  a),  (x  —  b), 
and  (a;  —  c) ;  three  of  the  second  degree,  (a;  —  a)  (x  —  &),  (x  —  a) 
(x  —  c),  and  (x  —  h)  (x  —  c) ;  and  one  of  the  third,  (x  —  a)  (x  —  b) 
(x  —  c).  These  divisors  are  evidently  equal  to  the  number  of  combi- 
nations which  can  be  formed  by  combining  3  letters  in  sets  of  1  and  1, 
2  and  2,  3  and  3.  So,  likewise,  if  we  take  the  m  factors  of  an  equation 
and  multiply  them  two  and  two,  three  and  three,  &c.,  we  shall  evidently 
obtain  as  many  divisors  of  the  second  degree  as  we  can  form  combina- 
tions of  VI  letters,  taken  two  and  two ;  and  as  many  divisors  of  the  third 
degree  as  there  are  combinations  of  ??i  letters  taken  three  and  three. 


440 


GENERAL  THEORY  OF  EQUATIONS. 


&c.     The  given  equation  will  then  have  m  divisors  of  the  first  degree, 

m(m  — 1)                                                         m(m  —  1)  (m  —  2) 
in  X,  ■ — — - —   divisors  oi   the  second  degree,  — ^= ^ — ^-^ 

of  the  third  degree,  &c. 


Fourth  Property. 

451.  All  the  coefficients,  after  the  first,  of  an  equation  of  the  m*'' 
degree,  are  functions  of  the  values. 

Suppose  the  general  equation  of  the  ?n*''  degree,  a;"  +  Pa:;"'-'  -f 
Qaj"'-^  +  .,..+  Tec  +  U  =  0,  contains  the  m  values,  ±  a,  d=  6,  ±  c, 
±  c7,  &c.  The  equation  can  then  be  put  under  the  form  of  (x  rp  a) 
(x  =F  6)  (x  =F  c)  (x  =fi  d),  &c.,  to  m  factors  =  0.  By  actually  per- 
forming the  indicated  multiplication,  we  will  get. 


=Fa 

x»-'  X  ah 

=F& 

Xac 

qpc 

Xad 

z^d 

t&c 

&c. 

Xhc 

&c. 

'  zp  aJc 
qr  ahd 
zp  ac(^ 

=p  &c. 


x""-^ =p  ahcd,  &c.  =  0. 


The  upper  row  of  signs  belonging  to  the  positive  value,  and  the 
lower  to  the  negative. 

We  observe  the  following  relation  between  the  coefficients  and  the 
values : 

1.  The  coefficient  of  the  second  term  is  the  sum  of  all  the  values  of 
the  unknown  quantity,  with  their  signs  changed. 

2.  The  coefficient  of  the  third  term  is  the  sum  of  the  products  of  all 
the  values,  taken  two  and  two,  with  their  respective  signs. 

3.  The  coefficient  of  the  fourth  term  is  the  sum  of  the  products  of  all 
the  values,  taken  three  and  three,  with  their  signs  changed. 

4.  The  last  term  is  the  continued  product  of  all  the  values  of  the 
unknown  quantity,  with  their  respective  signs  if  the  degree  of  the 
equation  be  even,  or  with  their  signs  changed  if  it  be  odd. 

We  have  supposed  in  the  preceding  demonstration  that  all  the  values 
were  positive  or  all  negative.  But  the  law  of  formation  for  the  coeffi- 
cients is  evidently  the  same  when  some  of  the  values  are  positive  and 
some  nejrative. 


GENERAL  THEORY  OF  EQUATIONS.        441 

Thus,  let  tte  values  be  +  a,  +  h,  and  —  c. 


Then,  (x  —  a)  (x  —  h)  (x4-c)=0,  orx^  —  a  \x^-\-a'b 

—  5        —  ac 

+  c\      — he 


X  +  ahc  =  0. 


And  we  see  that  coefficients  are  formed  in  accordance  with  the  above 


452.  The  preceding  properties  show  several  important  things  in 
relation  to  the  composition  of  an  equation. 

1.  If  the  coefficient  of  the  second  term  of  an  equation  be  zero,  and, 
consequently,  the  second  term  be  wanting,  the  sum  of  the  positive 
values  must  be  equal  to  the  sum  of  the  negative  values. 

2.  If  the  signs  of  the  terms  of  an  equation  be  all  positive,  the  values 
must  be  all  negative.  For,  an  equation  cannot  have  all  its  terms  posi- 
tive unless  its  binomial  factors  are  of  the  form,  (x  +  a)  (x  +  h)  {x  +  c), 
&c. ;  which  factors,  placed  equal  to  zero,  will  give  the  negative  values, 
—  a,  —  h,  —  c,  &c. 

3.  If  the  signs  of  the  terms  of  an  equation  be  alternately  +  and  — , 
the  values  of  the  unknown  quantity  will  be  all  positive.  For,  in  this 
case,  the  first  member  of  the  equation  must  be  made  up  of  the  factors, 
(x  —  a)  (x  —  U)  (x  —  c),  &c.,  corresponding  to  positive  values. 

4.  Since  the  last  term,  irrespective  of  its  sign,  is  the  continued  pro- 
duct of  the  values,  it  follows  that,  when  the  last  term  is  zero,  one  value, 
at  least,  must  be  zero. 

5.  It  follows,  also,  from  the  composition  of  the  last  term,  that,  in 
seeking  for  a  value,  we  need  only  seek  among  the  divisors  of  the  last 
term,  since  every  value  must  be  a  divisor  of  that  term. 

6.  If  we  know  one  value,  the  coefficient  of  the  second  term  will  give 
the  sum  of  all  the  rest,  and  the  last  term,  divided  by  this  value,  with  or 
without  its  sign  changed,  will  give  the  product  of  all  the  rest. 


Corollary. 

453.  1st.  The  last  two  consequences  enable  us  to  solve  equations 
with  facility  when  all  the  values  except  two  are  known. 

Take,  as  an  example,  the  equation,  x^  —  Ix'^  —  |x  +  |  =  0. 

The  values  must  be  sought  among  the  divisors  of  the  last  term : 
+  1  is  a  divisor  of  the  last  term,  and  may,  therefore,  be  a  value ;  upon 


442  GENERAL    THEORY    OP    EQUATIONS. 

trial,  we  find  that  it  will  satisfy  the  equation,  and  is,  therefore,  a  value. 
Calling  the  other  two  values  x  and  y,  we  will  have,  —  (1  -\-  x  -}- 1/)  = 

—  I,  the  coefficient  of  the  second  term.     Hence,  x  +  y  =  —  ^.     We 

xy 
have,  also  (sixth  consequence),  -f  =  —  i-      Combining   these   two 

equations,  we  get  x  ( —  i  —  a;)  =  —  ^,  or  cc^  +  |x  =  +  |.  And 
the  values  of  x  and  ^  are  found  to  be  +  ^,  and  —  1.  The  equation 
has,  of  course,  but  three  values,  since  it  is  an  equation  of  the  third 
degree,  (Art.  450). 

2d.  When  all  the  values  are  known  except  two,  we  can  tell  whether 
these  values  are  real  or  imaginary  without  actually  finding  them. 

Thus,  take  the  equation,  x*  —  6x^  —  IQx  +  21  =  0. 

The  exact  divisors  of  the  last  term  are  +1,  +3,  and  +7,  —  1, 

—  3,  and  —  7 ;  the  values  must  be  sought  among  these  numbers. 
We  find,  upon  trial,  that  +  1  and  +  3  are  values.  The  second  term 
of  the  equation  being  wanting,  the  sum  of  the  other  two  values  must 

21 

be  —  4,  and  their  product  :j — -  =  7.      But  4  is  the  greatest  product 

which  can  be  given  by  two  numbers  whose  sum  is  4.  Hence,  the 
other  two  values  of  the  equation  are  imaginary.  And,  in  fact,  by  pur- 
suing the  preceding  process,  we  find  them  to  be  — 2  -\-y/ — 3,  and 

_  2— ^ITS- 


EXAMPLES. 

1.  Solve  the  equation,  x^  —  2ax^  —  da'x  +  ISa^  =  0. 

Ans.   4-  2a  -f  3a,  and  —  3a. 

2.  Solve  the  equation,  x"  —  4x^  +  x  —  4  =  0. 

Ans.   +  4,  +  v/ —  1,  and  — ^ —  1. 

3.  Solve  the  equation,  x*  —  2x'  —  S  =  0. 

Ans.  +  2,  —  2,  +  ^/■=^,  —^':^^. 

454.  3d.  If  the  values  are  known,  the  equations  which  give  those 
values  can  be  formed  in  two  ways,  either  by  multiplying  together  the 
binomial  factors  corresponding  to  those  values  (Art.  450),  or  by  form- 
ing the  coefficients,  (Art.  451). 


GENERAL    THEORY    OF    EQUATIONS.  443 


EXAMPLES. 

1.  Form  tte  equation  whose  values  are  +  1,  —  1  and  —  2. 

Ans.  x''  +  2x^  —  x  —  2  =  0. 

For  the  factors,  multiplied  together,  and  placed  equal  to  zero,  give 
(x  —  1)  (x  +  1)  (a;  +  2)  =  0     (A)  ov  x'' +  2x''  —  x  —  2  =^  0. 

We  see  that  the  three  factors  of  (A),  placed  separately  equal  to  zero, 
will  give  the  values,  +  1,  —  1,  and  —  2,  or  the  problem  may  be  solved 
by  Art.  450. 
Coefficient  of  2d  term  =  —  (+1  —  1  —  2)=  +  2 

3d  term  =  (+1)  (-l)+(+l)  (-2)+(-l)  (-2)=-l 
«  4thterm=— (4-  1)  (— 1)  (— 2) =  —  2. 

2.  Form,  by  both  methods,  the  equation  whose  values  are  +  4, 
+  ^^^,  and  —  v/—  1.  Tins.  ar'  —  4x^+x  — 4  =  0. 

3.  Form,  by  both  methods,  the  equations  whose  values  are  +  2,  — 2, 
+  ^^-2,  —  ^^^.  Ans.  x*  —  2x^  —  S  =  0. 

4.  Form,  by  both  methods,  the  equation  whose  values  are  —  1,  —  2, 
—  3,  and  —  4.  Ans.  x*  +  lOx''  +  35x2  +  50x  +  24  =  0. 

Fifth  Property. 

455.  Every  equation  may  be  transformed  into  another,  in  which  the 
values  of  the  unknown  quantity  shall  be  equal  to  those  of  the  proposed 
equation,  increased  or  diminished  by  a  certain  quantity. 

Let  the  given  equation  be 

'    x"  +  Px»-'  +  Qx"'  + +  Tx  +  U  =  0, 

and  let  it  be  proposed  to  transform  it  into  another  in  y,  so  that  y  = 
x±  a.  The  principle  of  transformation  is,  of  course,  the  same  when 
y  =  x  +  o,  or  =  x  —  a  ;  we  will  then  confine  our  discussion  to  the 
case,  when  y  =■  x  —  a.  From  this  equation  there  results,  x  —  y  +  <^- 
Substituting  this  value  in  the  proposed  equation,  it  becomes  {jy  +  a)"* 

+  P  (^  +  a)-'  +  Q(y  +  ay-^  + T  (//  +  a)  +  U  =  0. 

Developing  the  different  powers  of  y  +  a  by  the  binomial  fonnula,  and 
arranging  the  development  according  to  the  descending  powers  of  y,  we 
have  the  transformed  equation, 


44 

GENERAL    THEORY 

OF    EQUATIONS. 

2/°  +  mo 
+  P 

„_,       in(m  —  1)  a^ 
^                   1.2 

+  «!,  (m  —  1)  Pa 

+  Q 

y"'-^  .   .   .   .    +  a" 

+  Pa'°- 
+  Qan-= 
4-Ta 

+  u 

=  0. 


If  the  values  of  y  in  this  equation  can  be  found,  the  corresponding 
values  of  x  in  the  given  equation  will  be  equal  to  these  values  in- 
creased by  a. 


Corollary. 

456.  The  preceding  transformation  enables  us  to  reduce  the  number 
of  terms  in  an  equation.  For,  since  the  quantity,  a,  is  entirely  arbi- 
trary, such  a  value  may  be  given  to  it  as  will  reduce  the  coefficient  of 
any  term  in  the  transformed  equation  to  zero,  and  consequently  make 
the  term  itself  disappear.  Suppose  we  wish  to  free  the  transformed 
equation  of  its  second  term;  we  have  only  to  place  the  coefficient  of  that 
term  equal  to  zero,  and  find  the  value  of  the  quantity,  a,  that  makes  it 
zero.  Of  course,  then,  this  value  attributed  to  a,  will  cause  the  dis- 
appearance of  the  second  term  of  the  equation.     Placing  ma  -f  P  =  0, 

P  P 

we  get  a  = .     Then,  x  i=  y .     Hence,  for  transforming  one 

equation  into  another,  in  which  the  second  term  is  wanting,  the  following 


RULE. 

Substitute  for  the  unknown  quantity  a  new  unknown  quantity,  con- 
nected with  the  quotient  arising  from  dividing  the  coeffi/^ient  of  the 
second  term  of  the  given  equation,  with  its  sign  changed  hy  the  degree 
of  the  equation. 


EXAMPLES. 

1.  Transform  the  equation,  x^  —  ^x  =  5,  into  another,  in  which  the 
second  term  shall  be  wanting.  Ans.  x  =  y  —  ( —  |)  =  y  +  2. 

The  transformed  equation  will  then  be  {y  -\-Tf  —  4  (y  -f  2)  =  5,  or 
y  —  4  =  5.  Hence,  y'^  =  9,  and  y  =  =1=  3 ;  and,  cc  =  y  -j-  2  =  5, 
or  —  1 ;  the  same  result  that  we  would  obtain  by  solving  the  given 
equation. 


GENERAL  THEORY  OF  EQUATIONS.         445 

2.  Transform  the  equation,  x^  —  Zx^  =  —  2,  into  another,  in  •which 
the  second  term  shall  be  wanting.  Ans.  y^  —  3y  =  0. 

The  transformed  equation  can  be  readily  solved,  one  value  of  y  being 
zero.  Hence,  one  value  of  the  given  equation  is  1.  The  transformed 
equation,  in  this  case  and  in  many  others,  is  simpler  than  the  given 
equation,  and,  therefore,  more  readily  solved.  The  chief  object  of  the 
transformation  is  to  simplify  the  form  of  the  equation. 


Scholium. 

457.  1.  The  third  term  of  the  transformed  equation  may  be  made  to 
disappear  by  giving  to  a  such  a  value  as  will  satisfy  the  equation, 

— ~ ~ —     +    (m  —  1)  Pa  +  Q  =  0.     And,  since  this  equation 

is  of  the  second  degree  in  a,  there  are  two  values  for  a,  and,  con- 
sequently, the  third  term  can  be  made  to  disappear  in  two  ways.  In 
like  manner,  the  fourth  term  can  be  made  to  disappear  in  three  ways, 
the  fifth  term  in  four  ways,  &c.  j  and  the  last,  or  {in  +  l)**"  term,  in  m 
ways.  By  recurrence  to  the  derivations  of  these  coefficients  from  the 
values,  we  see  that  the  above  results  are  true.  The  last  term,  for  in- 
stance, being  made  up  of  the  product  of  the  m  values,  can  be  made  to 
disappear  by  placing  either  of  the  m  values  equal  to  zero. 

458.  2.  It  may  happen  that  the  value  of  a,  which  makes  the  second 
term  disappear,  will  also  cause  the  disappearance  of  the  third  or  some 
other  term  at  the  same  time.  Let  us  examine  under  what  circumstan- 
ces the  second  and  third  terms  will  disappear  together.  By  placing  the 
coefficient  of  the  second  term,  Pa  +  m,  equal  to  zero,  we  get  a  = 

,  and,  by  placing  that  of  the  third  term,      ^    — ^;~—  -f  (m  —  1) 

P^       /P  2Q 


P  /P^ 

Pa  +  Q,  also  equal  to  zero,  we  get  «  ^ d=  \  /  — 

m       \   m        m(vi  —  1) 

Now,  it  is  plain  that  the  values  of  a  in  this  equation  will  be  identical 
P 
with  the  last,  that  is,  both ,  when  the  radical  disappears.     Placing 

2mQ 

the  radical  equal  to  zero,  we  get  P^  ^  — -.     "Whenever,  then,  the 

m  —  1 

square  of  the  coefficient  of  the  second  term  is  equal  to  twice  the  quo- 
tient arising  from  dividing  the  product  of  the  desree  of  the  equation 
38 


446         GENERAL  THEORY  OF  EQUATIONS. 

into  tte  coefficient  of  the  third  term  by  the  degree  of  the  equation, 
less  one,  the  second  and  third  terms  will  disappear  together. 

The  same  truth  may  be  demonstrated  more  elegantly  by  the  principle 
of  the  greatest  common  divisor. 

Dividing  the  coefficient  of  the  third  term  by  that  of  the  second,  we 

get,  after  two  divisions,  a  remainder,  mQ  +  ^^ ^ •  And,  it  is  evi- 
dent that,  when  this  remainder  is  placed  equal  to  zero,  we  will  have 
made  manifest  the  circumstances  under  which  there  will  be  a  common 
factor  between  the  two  coefficients.     Placing  the  remainder  equal  to 


zero,  we  get  P'^  = zr,  as  before 


EXAMPLES. 

1.  Make  the  second  and  third  terms  disappear  from  the  equation, 
x^  —  Bx^  +  Sx  —  28  =  0,  and  find  one  value  of  x. 

Ans.  Transformed  equation,  ^^  —  27  =  0,  and  a;  =  4. 

2.  Find  two  values  in  the  equation,  x*  -f  4a;'  -f  6x^  -\-4x  — 15  =  0. 
Ans.  Transformed  equation,  i/*  —  16  =  0,  then,  y  =  ±  2,  and  x  = 

+  1,  or  —  3. 

Sixth  Property. 

459.  Every  equation  having  the  coefficient  of  the  first  term  plus 
unity,  and  the  other  coefficients  entire,  will  have  entire  numbers  only 
for  its  rational  values. 

Let  the  proposed  equation  be 

^m   +  p^m-1   _|_  Qa.m-2 +   Tx  +  U  =  0. 

In  which,  P,  Q,  &c.,  are  whole  numbers. 

If  X  can  have  a  fractional  value  in  this  equation,  let  this  value,  re- 
duced to  its  lowest  terms,  be  — .  Substituting  this  value  for  x  in  the 
given  equation,  it  becomes 

Multiplying  by  h^~\  and  transposing,  we  get 

^  =  —  Pa"-'  —  Qa^-^i —  TaS^-^  —  U6"-». 

0 


GENERAL  THEORY  OF  EQUATIONS.         447 

But  tlie  second  member  is  an  entire  quantity,  since  all  its  terms  are 
entire.  Hence,  the  first  member  must  be  entire.  But,  since  a,  by 
hypothesis,  is  prime  with  respect  to  &,  a"  must  also  be  prime  with  re- 
spect to  b.  The  supposition  of  the  proposed  equation  containing  a  ra- 
tional fractional  value,  has  then  resulted  in  the  absurdity  of  making  an 
entire  quantity  equal  to  an  irreducible  fraction.  "We  conclude,  there- 
fore, that  this  supposition  is  wrong,  and  that  the  rational  values  are  all 
entire.     The  demonstration  is  restricted  to  rational  values,  because  the 

assumed  value,  -^,  is  rational  in  its  form. 


CoroEari/. 

460.  Articles  452,  456  and  459,  are  used  in  solving  numerical  and 
literal  equations  by  changing  their  forms. 

Let  the  proposed  equation  be  y*  —  -ii/^ —  8?/  +  32  =  0.  "We  know 
that  all  the  rational  values  are  entire  (Art.  459),  and  that  they  must  be 
found  among  the  divisors  of  the  last  term  (Art.  452).  But  as  there 
are  so  many  divisors  of  the  last  term,  it  will  be  more  convenient  to 
employ  (Art.  456)  to  transform  the  equation  into  another,  whose  last 

term  admits  of  fewer  divisors.     Make  re  =y  —  (—j-\  =  i/-\-l.   The 

transformed  equation  in  x,  is  x*  —  6.7=^  —  IGx  +  21  :=  0,  and  the 
divisors  of  the  last  term,  are  -}-  1,  +  3,  +  7,  +  21,  and  —  1,  —  3, 
—  7,  —  21.  On  trial  of  these  divisors,  wc  find  that  -f  1  and  +  3, 
will  satisfy  the  transformed  equation,  and,  consequently,  are  values. 
Hence,  cc  =  2/  +  1  =  2  and  4,  in  the  given  equation.  Dividing  the 
first  member  of  the  given  equation  by  (x  —  2)  (Art.  449),  it  will  be 
reduced  to  a  cubic  equation,  and  again  dividing  by  (x  —  4),  it  will  be 
reduced  to  a  quadratic,  which  can  be  solved. 

Seventh  Projierii/. 

461.  Imaginary  values  enter  equations  hi/  pairs. 

We  are  to  show  that  if  a-\-hy/ —  1  is  a  value  in  the  equation,  x"  -f 

Pa;"-'  +  Qx'"-^ +  Tx  +  U  =  0,  a  —  h^'^^  will  also  be  a 

value. 

The  truth  of  the  proposition  is  an  evident  consequence  of  Art.  451, 
for  U,  the  last  term,  is  the  product  of  all  the  values,  and  it  has  been 
assumed  real.     But  it  can  only  be  real,  when  the  imaginary  values  (if 


448         GENERAL  THEORY  OF  EQUATIONS, 

any),  enter  by  pairs,  and  of  such  a  form  as  to  give  a  real  product  when 
multiplied  together.  Thus,  if  we  have  two  imaginary  values,  m  + 
v/ —  n  and  ^ —  4,  we  must  also  have  two  other  values,  m  —  ^ —  n 
and  — v/ —  4,  otherwise,  U  will  not  be  real.  Or,  we  may  demonstrate 
the  property  otherwise,  thus :  by  substituting  the  assumed  value, 
a  +  h^ —  1,  in  the  given  equation,  we  will  have,  (a  -f  l-y —  1)"  +  P 

(a  +  i^/—  I)"-'  + T  (a  +  hs/^^)  +  U  =  0.      When 

these  terms  are  expanded,  the  odd  powers  of  t  v' —  1  will  be  imaginary, 
and  the  even,  real.  Representing  by  A  all  the  real  terms  involving  a 
or  h,  and  by  B^-^  1  aU  the  imaginary  terms,  we  will  have  A  + 
B  v' —  1  =  0.  Now,  since  a  +  h^ —  1  is,  by  hypothesis,  a  value,  the 
last  equation  must  be  satisfied,  but  this  can  only  be  so  when  A  =  0,  and 
Bv/ —  1  =  0,  since  imaginary  terms  cannot  be  cancelled  by  real  ones. 
If  «  —  h^ —  1  be  substituted  in  the  equation,  the  expanded  results 
will  be  precisely  the  same,  except  that  the  odd  powers  of  hy/ —  1  will 
be  negative.  Hence,  we  will  have  A  —  Bv^ —  1  =  0.  But  in  order 
that  a  -f  h^ —  1  should  be  a  value,  we  have  seen  that  A  =  0  and 
B  =  0.  Hence,  the  equation,  A  —  B  v' —  1  =;  0,  is  satisfied,  and  that 
being  so,  a  —  h-y —  1  must  be  a  value. 

The  absurdity  of  the  hypothesis  of  a  single  imaginary  value  may  be 
illustrated  by  an  example. 

Let  us  assume  that  the  three  values  of  an  equation  of  the  third 
degree  are  +  a,  +1,  and  +  v/ — 1.  The  equation  must  then  be 
(x  —  a)Xx  —  1)  (x  — vA=n:)  =  0,  or  (x  —  a)  (x"  —  (1  — v/=l) 
35  + >/ — 1)  =  0.  Hence,  a?  —  a  =  0,  andcc''  —  (l  +  \/ — V)x-\-y/ — 1 

=  0.     The  second  equation,  when  solved,  will  give  x  = ~ 


\/-^..-r+  i.2^-i-i,L±v^i^^/. 


!^/-l 


2 

And  we  see  that  we  do  not  get  back  again  the  two  values,  +  1,  and 
+V-1. 

Remarks. 

462.  1.  The  product  of  imaginary  values  is  always  positive,  and, 
therefore,  the  absolute  term  of  an  equation,  whose  values  are  all  imao-i- 
nary,  must  be  positive. 

2.  If  the  second  term  of  an  equation  containing  only  imao-inary 
values,  is  wanting,  these  values  will  all  be  of  the  form,  ±v'— ^• 

3.  Every  equation  of  an  odd  degree  has  at  least  one  .real  value  forr, 


GENERAL  THEORY  OP  EQUATIONS.         449 

and  the  sign  of  this  value  will  be  contrary  to  that  of  the  last  term  of 
the  equation. 

4.  Every  equation  of  an  even  degree,  whose  last  term  is  negative, 
has  at  least  two  real  values  for  x  ;  one  positive,  and  one  negative. 


EXAMPLES. 


1.  One  value  of  cc  is  4  +  ^ —  10,  what  must  a  second  value  be? 

Anz.  4— ^=IOr 

2.  Two  of  the  values  of  an  equation  of  the  fourth  degree  are  +  a, 
and  —  hj  and  the  last  term  is  —  m.     How  are  the  remaining  values  ? 

Ans.  Real. 

Eighth  Property. 

463.  If  the  real  values  of  an  equation,  taken  in  the  order  of  their 
magnitudes,  be  n,  h,  c,  d  and  e  ;  a  being  ^6,  h'^  c,  &c. ;  then,  if  a 
series  of  quantities,  a',  h',  c',  d',  &c. ;  a'  taken  greater  than  a,  V  be- 
tween a  and  h,  c'  between  h  and  c ;  be  substituted  for  x  in  the  pro- 
posed equation,  the  results  will  be  alternately  positive  and  negative. 

For,  since  a,  &,  c,  d,  &c.,  are  assumed  to  be  values,  the  proposed 
equation  can  be  put  under  the  form  of  (x — a)  (x  —  h)  (x  —  c)  (x  —  d) 
&c.  =  0. 

Substituting,  for  x,  the  proposed  series  of  quantities,  a',  h',  d,  d',  &c., 
we  get  the  following  results. 

(of  —  a)  (a'  —  b)  (a'  —  c)  (a'  —  d)  &c,  =  -{-  result,  since  all  the 
factors  are  positive. 

(6'  —  a)  (b'  —  I)  (V  —  r)  (U  —  d)  kc,  =  —  result,  since  first  factor 
only  is  negative. 

(c'  —  a)  (d  —  h)  (d  —  c)  (c'  —  d)  &,c.,  =  +  result,  since  first  two 
factors  only  are  negative. 

(df  —  a)  ((Z'  —  5)  (d'  —  c)  (d'  —  d)  &c.,  =  —  result,  since  first 
three  factors  only  are  negative 

Remarks. 

1.  We  see  that,  between  the  quantities,  a'  and  h',  which  give  the  first 
plus  result  and  the  first  minus  result,  there  is  one  value,  a.     And  be- 
tween the  quantities,  a'  and  d',  which  give  the  first  plus  result  and  the 
second  minus  result,  there  are  three  values,  a,  h  and  c.     And,  as  the 
38*  2d 


450         GENERAL  THEORY  OF  EQUATIONS. 

same  is  evidently  true  for  any  number  of  odd  values,  we  conclude  that, 
if  two  quantities  be  successively  substituted  for  x  in  an  equation,  and 
give  results,  with  different  signs,  between  these  quantities,  there  will 
be  one,  three,  five,  or  some  odd  numbers  of  real  value. 

2.  If  any  quantity,  m,  and  every  quantity  greater  than  m,  give  re- 
sults all  positive,  then  m  is  greater  than  the  greatest  value  of  x  in  the 
equation. 

3.  If  the  results  obtained  by  substituting  two  quantities  have  the 
same  sign,  then,  between  these  quantities  there  are  two,  four,  six,  or 
some  even  number  of  values,  or  no  value  at  all.  Thus,  between  a'  and 
c',  which  give  +  results,  there  are  two  values,  a  and  h. 

Scliolium. 

464.  The  preceding  property  may  be  demonstrated  differently,  by 
employing  a  principle  of  frequent  application  in  all  branches  of  mathe- 
matics, viz.,  that  a  quantity  changes  its  sign  in  passing  through  zero 
and  infinity.  Let  the  quantity  be  x  =  a  ;  suppose  x  ^  a,  the  expres- 
sion is  positive ;  it  is  zero  when  x  =  a,  and  negative  when  x<^a. 
Hence,  the  quantity  x  —  «,  changes  its  sign  in  passing  through  zero. 

+  M 
Again,  take ,  This  quantity  is  positive  when  x'^a  ;  it  is  in- 
finite when  X  =  a ;  and  negative  when  x  <^a.  Now,  in  the  results 
of  Art.  463,  between  the  first  +  result  and  the  first  —  result,  the  first 
member  of  the  proposed  equation  must  have  passed  through  zero,  and 
consequently  through  a  value.  In  like  manner,  between  the  first 
—  result,  and  the  second  +  result,  the  first  member  must  again  have 
passed  through  zero,  and  consequently  through  a  second  value.  Then 
two  values  must  have  been  passed  through  between  the  first  two  +  re- 
sults, &c.  By  continuing  the  same  course  of  reasoning,  we  might 
readily  show  that  three  values  must  be  passed  through  between  the  first 
positive  and  second  negative  result;  five  values  between  the  first  posi- 
tive and  third  negative  result;  two  values  between  the  first  two  negative 
results;    four  between  the  first  and  third  negative  results,  &c.,  &c., 

Limit  of  the  Values.  —  Ninth  Property. 

465.  If  we  substitute  for  the  unknown  quantity  the  greatest  coeffi- 
cient plus  unity,  the  first  term  of  the  equation  will  be  greater  than  the 
sum  of  all  the  other  terms,  provided  we  always  affect  the  greatest  co- 
efficient with  the  positive  sign. 


GENERAL  THEORY  OF  EQUATIONS.         -151 

Let  us  resume  the  equation, 

x^  +  Prr-"-'  +  Qx"-^ +  Ta;  +  U  =  0. 

It  is  plain  that  the  most  unfavourable  case  will  be  when  all  the  co- 
efficients have  the  same  sign,  -f  or  — .  We  will  assume  all  the  coeffi- 
cients after  the  first  to  be  negative ;  the  given  equation  will  then  be  of 

the  form,  x'^  —  Tx^-'  —  Px"-^  —  ¥x'^-^—  P:c»-^ —  P  =  0, 

or  a;"  _  P(a;"-'  —  x""''  —  x"'"^ —  1)  =  0. 

Disregarding  the  sign  of  P  for  a  moment,  we  wish  to  ascertain  what 
value  X  must  have,  in  order  that  cc"  ^  P(x°~'  -f-  cc"^^  +  a;"~*  .  .  .  -f  x 
+  1).  Inverting  the  order  of  the  terms  within  the  parenthesis,  we 
will  have  a  geometrical  progression,  whereof,  1  is  the  first  term,  x°^' 
the  last  term,  and  x  the  ratio  of  the  progression.     The  formula,  S'  = 

— ,  will  then  give  the  value  of  the  quantity  within  the  parenthesis,  ^ 

a;"  — 1      _  ,  ^    ^    /:,•■»— K  ^       Px"  P 


(7-m   |v  rX 


^_^.  ....... ,..^.....  ^.^^_^^,...  ^x  —  1    x  —  r 

Px" 
This  inequality  will  evidently  be  true,  when  x"  = -.      Dividing 

both  members  of  this  equation  by  a;",  and  clearing  of  fractions,  we  find 
a;  =  P  +  1,  as  enunciated. 

It  is  to  be  observed  that,  in  this  demonstration,  we  assume  a;  ^  1. 

EXAMPLES, 

1.  What  number  substituted  for  x  in  the  equation,  x^  —  7x*  -f  Gx'^  -f 
6x^  -f  X  =  0,  will  make  the  first  term  greater  than  the  sum  of  all  the 
other  terms  ?  A71S.  8. 

2.  What  in  the  equation,  x*—  lOa;^  +  12x  +  13  =  0  ? 

Ans.  14. 

3.  What  quantity  in  the  equation,  x^ —  2aa;^  -f-  6ax  —  12a  =  0. 

Ans.  1  4-  12a. 

JScJiolium. 

In  seeking  for  the  positive  values  of  an  equation,  we  need  not  go 
beyond  the  greatest  coefficient  with  a  positive  sign  prefixed  to  it,  plus 
unity.  Thus,  in  example  first,  there  is  no  positive  value  greater  than 
8,  because  8,  and  all  numbers  greater  than  8,  will  continually  make  the 
first  member  positive  (Art.  463). 


452         GENERAL  THEORY  OF  EQUATIONS, 


Second  Limit.  —  Tenth  Property. 

466.  If  we  substitute  for  x  in  the  equation,  x"  +  Pa:"-'  +  Qx"-^ 
.  .  .  ,  Ta;  +  U  =  0,  unity,  increased  by  that  root  of  the  greatest  nega- 
tive coefficient  which  is  indicated  by  the  number  of  terms  preceding 
the  first  negative  term,  we  will  have  a  superior  limit  of  the  positive 
values. 

Suppose  Nx""""  to  be  the  first  negative  term,  and  suppose  W  to  be 
the  greatest  negative  coefficient.  The  most  unfavourable  case  obviously 
is  that  in  which  we  suppose  all  the  coefficients  after  N,  negative,  and 
equal  to  W.     It  is  plain,  moreover,  that  any  value  of  x  which  makes 

a."  ^  W(x'"-°  +  x"-"-' +  «+!),  will,  of  course,  make  the 

first  member  of  the  given  equation  positive.  We  are  to  find  the  value 
which  fulfils  this  condition. 

Resuming  the  inequality,   x"  ^  W(a;"-°  +  a"-"-'  +  ic""-""^  .... 

^   TTT  /*"-"+'  —  1\  ^    W£c"-'+' 

-j-  X  +  1),  we  have,  also,  x"  >  W  ( z — ),  or  x"  > -— 

'  "^        \     X  —  1/  -^x  —  1 

W  ....  'Wx'"~°+' 
-.     And  this  inequality  will  be  true,  when  x"  = 


1 

W 

From  which,  x  —  1  =  —^^ ,  or  (x  —  l)x"~'  =  W.     Now,  if  (x  —  1) 

(x  —  1)°-'  =  W,  then  {x  —  l)x"-'  will  be  greater  than  W.  If,  how- 
ever, we  place  it  equal  to  W,  we  will  have  (a;  —  1)  (x  —  1)"-'  =  W, 
or  (x  —  1)°  =  W.  Hence,  x  =  1  +  y  W,  which  agrees  vsdth  the 
enunciation. 

It  will  be  seen  that  in  two  respects  we  have  taken  an  unfavorable 
case.  The  second  limit  t  en  may  considerably  exceed  the  greatest 
positive  value,  but  it  is  smaller  than  the  first  limit,  and,  therefore,  the 
most  used  in  practice. 


EXAMPLES. 

1.  Required  superior  positive  limit  of  the  values  in  the  equation, 
x«  -f  2x*  —  Sx"  +  7x=^  —  17x  -F  5  =  0.  Am.  1  -^JW. 

2.  Required  superior  positive  limit  of  the  values  in  the  equation, 
a;<  +  4x»  -J-  12x2—  5x  — 16  =  0.  Am.  1  +  yle: 


GENERAL  THEORY  OP  EQUATIONS.         453 


Eleventh  Property. 

467.  The  equation,  x""  +  Pa;'"-'+  Qa;'"-^  + +  Ta;  +  U  =  0, 

can  be  transformed  into  another  in  y,  or  some  other  variable,  such,  that 
the  values  of  the  new  unknown  quantity  shall  be  some  multiple  of 
those  of  the  old  unknown  quantity. 

For,  let  y  =  nx,  then  x  =  —.  Substituting  for  x,  wherever  it  occurs 
'in  the  given  equation,  its  value  in  terras  of  y,  we  have  --  +  P  ^^ri  + 
Q^V....T^^+  U  =  0,  or  y- +  7iFy-i  +  n'Qy-' 


w^-'Ty  +  n"!!  =  0.  And  we  sec  that  the  transformation  is  effected 
by  changing  x  into  y,  multiplying  the  coefficient  of  the  second  term  by 
the  multiple  n,  that  of  the  third  term  by  n^,  &c.  For  instance,  let  it 
be  required  to  transform  the  equation,  x*  +  2x'  +  4x^  +  8ar  -f-  1  =  0, 
into  another,  in  which  the  values  of  the  new  unknown  quantity  shall 
be  twice  as  great  as  those  of  the  old.  Then  y  ^  2x,  and  we  have  y* 
+  4/+16/+64y  +  16  =  0. 

RemarJcs. 

The  principal  use  of  this  transformation  is  in  clearing  an  equation 
of  fractional  coefficients,  and  at  the  same  time  making  the  new  equa- 
tion preserve  the  proposed  form ;  that  is,  the  coefficient  of  the  first  term 

Pa;""' 
is  still  to  be  plus  unity.     Let  us  take  the  equation,  as™  -\ 1- 

Qx"-'   .   Kx"-'  Tx       U      ^  .       , .  , 

— ■  H ^  0,  in  which  we  assume  mn  to  be 

n  mn  n         n 

the  least  common  multiple  of  the  denominators.  Then  y  =  mnx. 
The  transformed  equation  will  then  be  ^-^^ — I h^— — ;-! ~ 5 

T»     m-3  Tl  TT 

m^b^Ry'"-^  ....  m"-'w"'-'^Ty  +  m^Ti^-'U  =  0.  And  we  see  that  the 
transformation  is  effected  by  changing  x  into  y,  and  multiplying  the 
second  term  by  the  first  power  of  the  least  common  multiple,  the  third 
term  by  the  second  power  of  that  multiple,  &c.     Thus,  the  transformed 

equation  of  x^  +  |-  —  |  +  1  =  0,  is  y'  +  4/  —  36y  +  1728  =  0. 


454  GENERAL    THEORY    OE    EQUATIONS. 

468.  The  two  rules  just  given  are,  of  course,  only  applicable  when 
the  coefficient  of  the  highest  power  of  x  is  plus  unity.  When  that  is 
not  the  case,  we  may  either  first  make  this  coefficient  plus  unity,  and 
apply  one  of  the  preceding  rules,  or  we  may  at  once  make  y  =  nx,  n 
in  this  case  representing  the  product  of  the  least  common  multiple  of 
the  denominators  by  the  coefficient  of  the  highest  power  of  x. 


EXAMPLES. 

1.  Transform  the  equation,  x^  —  4x'  +  2x  -f  2  =  0,  into  another, 
in  which  the  values  shall  be  twice  as  great  as  in  the  given  equation. 

Ans.  y  —  Stf  +  8y  +  16  =  0. 

2.  Transform  the  equation,  x*  —  ox'^  +  2a;^  +  a?  +  1  =  0,  into  an- 
other, whose  values  are  treble  those  of  x. 

A71S.  /  —  9if  -f  18/  +  27y  +  81  =  0. 

3.  Transform  the  equation,  x'^  —  "9-  +  -vr  +  k  +  1  =  0,  into  an 

other,  containing  only  entire  coefficients. 

Ans.  y  —  15/  +  300/  +  5400?/  +  810000  =  0. 

4.  Transform  the  equation,  2x^  —  "5"+ p  +  1  =  0,  into  another, 

o         O 

whose  coefficients  shall  all  be  entire. 

Ans.  2/  —  2/  4-  Gy  4-  216  =  0. 

5.  Transform  the  equation,  2x^  —  "o"  +  F  "^  1  =  0,  into  another, 

o        o 

whose   coefficients  shall  all  be  entire,  and  the  coefficient  of  the  first 
term,  plus  unity.  Ans.  /  —  2/  +  12y  +  864  =  0. 

In  this,  make  y  =  12a:;. 

469.  When  the  coefficient  of  the  first  term  is  some  whole  number 
different  from  unity,  and  the  other  coefficients  are  entire,  make  n,  in 
the  equation  y  =  nx,  equal  to  the  coefficient  of  the  first  term.  Thus,  to 
transform  the  equation,  Qx^  —  19^;^+  19x  —  6  =  0,  make  y  =  Qx, 
the  transformed  equation  will  be  /  —  19/  +  114y  —  216  =  0. 

Twelfth  Properti/.  —  Process  of  Divisors. 

470,  In  every  equation  in  which  the  coefficient  of  the  first  term  is 
unity,  and  all  the  other  coefficients  are  entire,  a  value  will  divide  the 
last  term,  the  sum  arising  from  adding  the  quotient  so  obtained  to  the 


GENERAL    TS'EORY    OF    EQUATIONS.  455 

coefficient  of  the  second  term  from  the  right,  the  sum  arising  from 
adding  this  second  quotient  to  the  coefficient  of  the  third  term  from  the 
right,  and  will  thus  continue  to  be  an  exact  divisor  of  all  the  successive 
sums  so  formed,  until  the  coefficient  of  the  second  term  from  the  left 
has  been  added  to  the  previous  quotient,  when  the  last  quotient  will  be 
minus  unity. 

Let  us  resume  the  equation, 

x^  +  Px"-'  +  Qx"'-^ +  Ta;  +  U  =  0, 

in  which  all  the  coefficients  are  entire.  Suppose  a  to  be  a  value,  the 
equation  will  then  become,  after  transposition  and  division  by  «,  a"~' 

+  Pa""-^  +  Qa°~^ T  = .     Now,  since  the  first  member  is 

made  up  of  entire  terms,  the  second  member  must  be  entire  also. 
Hence,  U  is  divisible  by  a,  as  it  ought  to  be,  since  U  is  the  product  of 
all  the  values.     Transposing  T  to  the  second  member,  we  have  a"—'  + 

P(j°'-2  ^  Qa'o-s  +  &c.  =  — T =  U';  U'  representing  the  alge- 
braic sum  of  T  and  — . 
a 

Dividing  both  members  again  by  a,  we  get  the  equation,  a"""  + 

Pa"-'  +  Qa"-^  +  &c.  = . 

a 

The  first  member  being  entire,  the  second  member  must  be  entire 
also.  Hence,  the  second  condition  to  be  fulfilled  by  a  value  is,  that  it 
must  be  an  exact  divisor  of  the  sum  formed  by  adding  the  quotient  of 
the  last  term  by  the  value  to  the  coefficient  of  the  first  power  of  x. 
This  was  also  to  have  been  anticipated,  because  the  coefficient  of  the 
first  power  of  x  is  made  up  of  the  values,  taken  m  —  1  and  m  -^  1 ; 
one  of  these  combinations  will  not  contain  the  value  a,  and  will  have  a 
sign  contrary  to  the  first  quotient,  and  will  be  cancelled  by  it.  The  re- 
maining terms  making  up  this  coefficient  all  contain  a,  hence,  a  will  be 
a  divisor.     By  transposing  the  terms  in  succession,  and  continuing  the 

.  .  .         P' 

division,  we  will  find,  after  m  transpositions  and  divisions,  —  =  —  1 ; 

P'  representing  the  sum  of  the  coefficients  of  x""-'  added  to  the  pre- 
ceding quotient. 

Hence,  the  last  condition  to  be  fulfilled  by  a  value  is,  that  it  must  be 
an  exact  divisor  of  the  sum  formed  by  adding  the  coefficient  of  x"-'  to 
the  preceding  quotient. 


456         GENERAL  THEORY  OF  EQUATIONS. 

We,  therefore,  have  the  following  rule  for  finding  the  rational  values 
of  an  equation. 

RULE. 

I.  Transform  the  equation,  if  necessary,  into  another,  in  lohich  all 
the  coefjicients  shall  he  entire;  that  of  the  first  term  heing  plus  unity. 
All  the  rational  values  will  then  he  entire  (Art.  459). 

II.  Find  the  superior  positive  and  superior  negative  limits  of  the 
values. 

III.  Write  down  in  succession,  in  the  same  horizontal  line,  all  the 
entire  divisors  of  the  last  term  hetween  zero  and  these  limits. 

IV.  Divide  the  last  term  hy  each  of  these  divisors,  and  write  the  re- 
spective quotients  hcneath  the  corresjyonding  divisors. 

V.  Add  the  coefiicient  of  the  first  power  of  x  to  each  of  these  quo- 
tients, and  tvrite  each  sum  thus  formed  heneath  the  corresponding 
quotient. 

VI.  Divide  these  sums  hy  the  corresponding  divisors  in  the  column 
of  divisors,  and  write  the  new  quotients  under  the  corresponding  sums, 
rejecting  those  divisors  lohich  give  fractional  quotients,  and  crossing  out 
the  vertical  column  in  which  they  occur. 

VII.  Proceed  in  this  way  until  the  coefficient  of  x""'  has  heen  added, 
to  the  preceding  quotient,  then  those  divisors  that  give  minus  unity  for 
quotients,  are  values,  and  they  are  the  only  rational  values. 

EXAMPLES. 

a;3  _  2a;2  —  5x  +  6  =  0. 

The  superior  positive  limit  is  1  +^5  =  6. 

The  superior  negative  limit  is  found  by  changing  +  x  into  —  x;  the 
superior  positive  limit  of  the  transformed  equation  will  then  be  the 
superior  negative  limit  of  the  given  equation. 

Changing  +  x  into  —  x,  we  get  —  x^  —  2x^  -f-  5a;  +  6  =  0.  But 
the  coefficient  of  the  first  term  must  always  be  plus  unity,  hence,  by 
multiplying  the  equation  by  minus  unity  we  have,  x^  +  2x^  —  bx  — 
6  =  0.  _ 

Then  —  (1  +  v'e)  =  superior  negative  limit.  Assume  —  4  to  be 
this  limit,  the  square  root  of  6  being  between  2  and  3,  we  take  3  to  be 
the  root,  since  it  is  better  to  have  the  hmit  too  great  than  too  small. 
Hence,  we  have  these  divisors  below  the  limits,  +1,  +  2,  +  3 ;  —  1, 
-2,-3. 


GENERAL    THEORY    OF 

EQUATIONS. 

Positire. 

Nesative. 

Divisors,     +  1,  +  2,  +  3, 

-    1,-2, 

Quotients,  +  6,  +  3,  +  2, 

-   6,-3, 

Sums,         +1,-2,-3, 

-11,-8, 

Quotients,  +1,-1,-1, 

+  11,  +  4, 

Sums,         —1,-3,-3, 

+    9,  +  2, 

Quotients,  —  1,     x ,  —  1- 

-   9,-1. 

457 


Positive. 

Divisors,     +    1, 

2,  +4, 

+  8, 

Quotients,  —    8, 

-4,-2, 

—  1, 

Sums,         —    8, 

-4,-2, 

—  1, 

Quotients,  —   8, 

—  2,      X, 

X, 

Sums,         — 10, 

—  4,      X, 

X, 

Quotients,  — 10, 

-2,      X, 

X, 

Sums,         — 10, 

—  2,      X, 

X  , 

Quotients,  — 10, 

—  1,        X, 

X, 

Hence,  we  have  +  1 ,  +  3,  and  —  2,  for  the  three  values  of  the 
given  equation. 

Let  us  take,  as  a  second  example,  the  equation,  x*  —  2x^  —  8  =  0. 
Superior  positive  limit,  =  +  9 ;  superior  negative  limit,  =  —  9. 
The  equation  may  be  written  x*  ±  Ox'*  —  2x^  ±0x  —  8  =  0. 


NegatiTe. 

—  1,-2,-4,-8, 
+  8,  +  4,  +  2,  +  1, 
+  8,  +4,  +2,  +1, 
-8,-2,       X,     X, 

—  10,-4,  X,  X, 
+  10,  +  2,  X  ,  X  , 
+  10,  +2,  X,  X, 

—  10,-1,  X,  X, 


Hence,  +  2  and  —  2  are  values  in  the  given  equation.  And  by 
dividing  the  first  member  of  the  given  equation  by  the  factors  (x  —  2) 
and  (x  +  2)  corresponding  to  these  values,  we  obtain  the  quotient, 
x^  +  2.  By  placing  this  quotient  equal  to  zero,  the  remaining  two 
values  of  the  equation  will  be  found  to  be  +  ^Z —  2,  and  — ^ —  2. 

471.  The  process  of  divisors  is  applicable  to  literal,  as  well  as  nu- 
merical equations,  when  all  the  coefficients  are  entire.  A  literal  equa- 
tion, too,  will  best  show  why  it  is  that  the  successive  sums  are  divisible 
by  the  values. 

Take,  as  an  example, 

x^  —  (a  +6  +  c)x'^  +  (o6  +  ac+  hc^x  —  ahc  =  0. 

Since  the  terms  are  alternately  plus  and  minus,  the  values  must  all 
be  positive. 
39 


458 


GENERAL  THEORY  OF  EQUATIONS. 


Divisors,     +  a,  -\-  b,  -{-  c. 
Quotients,  —  he,  —  oc,  —  ah. 
Sums,   ah  -j-  ac,  ah  +  he,  ac  +  he. 
Quotients,  h  -^  e,  a  -{-  c,  a  -\-  b. 
Sums,  —  a,  —  h,  —  c. 
Quotients,  —  1,  —  1,  —  1. 

Hence,  the  three  values  are  -\-  a,  +  h,  -\-  c. 

We  will  take,  as  a  second  example  in  literal  equations, 

a;3  4-  (a  —  6  -f  2')x^  +  {2(a  —  h')  —  ah)x  —  2al  =  0. 


Positive. 

Divisors,  +  o,  +  &,  +  2, 
Quotients,  —  2h,  —  2a,  —  ah, 
Sums,  (2a— ha  — 4&),  —  h(a  +  2), 

2(a  —  h  —  ah), 
Quotients,  X, — (a-\-2,)(a — h — ah}, 
Sums,  X,—h,  2(a+l)—h(a.  +  l), 
Quotients,  X  ,  —  1,  x. 


Negative. 

-  «,—  ^,—  2, 
+  2h,  +  2a,  +  ah, 
a(2  — &),4a  — &(2  +  a),2(a  — S), 


+  (h-2),X, 
+  a,  X  ,  +  2, 
—  1,  X,— 1. 


(a -6), 


Hence,  the  three  values  are  +  h,  —  a  and  —  2. 
The  process  of  divisors  being  of  such  high  practical  importance,  we 
will  give  another  example,  in  which  one  of  the  sums  is  zero. 
Take  the  equation, 

x''  —  x"  —ix  +  4  =  0. 
Superior  positive  limit,  =  5 ;  superior  negative  limit,  =  —  3. 


Tositive. 

Divisors,  +  1,  +  2,  +4, 
Quotients,  +  4,  -f  2,  +  1, 
Sums,  0,  —  2,  —  3, 

Quotients,  0,  —  1,  X , 
Sums,  — 1, — 2,  X, 
Quotients,  —  1,  —  1,      X  , 


Negative. 

-1,-2, 
-4,-2, 
-8,-6, 
+  8,  +  3, 
+  7,  +  2, 
-7,-1. 


Hence,  the  three  values  are  +  1,  +  2,  and  —  2. 

472.  The  process  of  divisors  enables  us  to  find  all  the  rational  values 
of  any  equation.  We  have  only  to  make  all  the  coefl&cients  entire,  if 
not  already  so,  that  of  the  highest  power  of  x  being  made  plus  unity. 
All  the  rational  values  of  y  in  the  transformed  equation  will  be  entire ; 
find  these  values,  and  then  those  of  x  will  be  known  from  the  relation 
between  x  and  y. 

Take  6x'  —  Idx^  +  19x  —  G  =  0.     Making  ^  =  Qx,  we  have  the 


GENERAL    THEORY    OF    EQUATIONS.  459 

transformed  equation  in  y,  ^  —  19j/^  +  114y  —  216  =  0.  By  pi*o- 
ceeding  as  before,  we  find  the  three  values  of  y  to  be  +  9,  +  6,  and 
+  4.     Hence,  the  corresponding  values  of  a;  are  |  +  1  and  +  |. 

473.  In  this,  and  in  many  examples,  the  last  term  has  a  great  num- 
ber of  exact  divisors,  and  the  process  is  therefore  tedious.  It  is  often 
convenient  to  transform  an  equation,  whose  last  term  is  too  large,  into 
another  whose  values  shall  be  greater  or  smaller  by  a  constant  quantity. 
Making  y  =  z  +  2,  in  the  preceding  equation,  the  transformed  equa- 
tion in  z  '\^  ^  —  132r^  +  50z  —  56  =  0 ;  and  the  three  values  of  z 
are  7,  4,  and  2.  Hence,  those  of  y  are  9,  6,  and  4,  and  those  of  x, 
-f  |,  -}-  1,  and  -f  I,  as  before. 

474.  If  we  solve  the  equation,  x^  —  2x''  —  3a;^  -f  8x  —  4  =  0,  by 
the  process  of  divisors,  we  will  get  the  three  values,  -f  2,  —  2,  and 
—  1.  And,  by  dividing  the  first  member  of  the  equation  by  the  factors 
corresponding  to  these  values,  we  will  have  left,  x  -f-  1  =  0.  Hence, 
the  value  —  1  enters  twice  in  the  given  equation.  We  see,  from  this 
example,  that  while  the  process  of  divisors  gives  the  rational  values,  it 
does  not  show  whether  any  or  all  of  them  are  repeated  once  or  more. 
This  remark  is  evidently  applicable  to  all  equations  whatever.  Some 
test  is  then  necessary,  by  which  we  can  ascertain  the  equal  values. 

EQUAL  VALUES. 

475.  Let  the  equation  be 

(x  — a)"  {x  —  hy  {x  —  cj  (x—d)  {x  —  e) &c.  =  0  (A), 

in  which  there  are  m  values  equal  to  a,  to  values  equal  to  h,  j^  values 
equal  to  c,  and  all  the  other  values  unequal.  Calling  D  the  differential 
coefficient  of  this  equation,  we  will  have  (Art.  366), 

D  =  m  (x  —  a)"-'  (x  —  bf  (x  —  c)p  (x  —  d)  (x  —  e)  &c.  + 
n  (x  —  6)"-'  (x  —  a)"  (x  —  cf  (x  —  d)  (x  —  e)  &c.  -\-p{x  —  c)p-' 
(x  — a)"  (x  — 6)„  (x— c?)(x  — c)&c.  +  (x— a)"'(x  — J)"(x  — c)P 
(x  —  e)  &c.  -f  (x  —  a)"  (x  —  5)°  (x  —  c)p  (x  —  d)  &c.,  plus  other 
terms  (B). 

It  is  plain  that  the  greatest  common  divisor,  D',  between  (A)  and 
(B),  will  be 

D'  =  (x  —  a)"-'  (x  —  &)"-'  (x  —  c)P-'. 
We  see  that  this  divisor  contains  (m  —  1)  values  equal  to  «,  (n  —  1) 
values  equal  to  h,  and  (p  —  1)  values  equal  to  c.     Hence,  it  is  plain 
that  the  number  of  equal  values  in  each  set  of  the  greatest  common 


4G0         GENERAL  THEORY  OF  EQUATIONS. 

divisor  is  one  less  than  in  the  giv      equation.     Therefore,  to  ascertain 
whether  there  are  equal  values,  we  have  this 


RULE. 

Find  the  differential  coefficient  of  the  first  memher  of  the  given  equa- 
tion [or,  as  it  is  generally  called,  the  first  derived  polynomial),  next  get 
the  greatest  common  divisor  hetiveen  the  first  member  of  the  pro^wsed 
equation  and  this  differential  coefficient,  place  this  common  divisor  eqtial 
to  zero,  and  find  its  values.  Each  independent  value  so  found  loill  he 
repeated  once  oftener  in  the  given  equation  than  in  the  greatest  common 
divisor ;  if  the  latter,  for  instance,  contain  two  values  equal  to  a,  and 
four  values  equal  to  b,  the  former  loill  contain  three  values  equal  to  a, 
and  five  equal  to  b.  If  the  greatest  common  divisor  he  of  too  high  a 
degree  to  solve,  we  may  find  the  second  differential  coefficient,  or  second 
derived  polynomial  of  the  given  equation,  then  the  greatest  common 
divisor  between  this  polynomial  and  the  first  memher  of  the  given  equa- 
tion. Each  independent  value  of  this  common  divisor  will  he  re- 
peated tioice  oftener  in  the  given  equation. 


EXAMPLES, 

1.  Does  the  equation,  x*  —  2x^  +  ^x^  —  x-\-h  =  0,  contain  equal 
values  ? 

First  derived  polynomial,  4x'  —  6a:^  +  3a;  —  1. 

Greatest  common-  divisor,  x  —  1,  which,  placed  equal  to  zero,  gives 
cc  =  l.  Hence,  th||  given  equation  contains  two  values  equal  to  1. 
If  we  divide  that  equatipn  by  (x  —  Vf,  we  will  have  left,  x^  +  l  =  Q. 
Hence,  the  other  two  values  are  +  v/ —  I,  and  — s/ —  J- 

2.  Does  the  equation,  x^  —  x^  —  14a;''  —  2Qx^  —  19a?  —  5  =  0,  con- 
tain equal  values  ? 

First  derived  polynomial,  5x^  —  Ax^  —  420;^  —  52x  — 19. 

Greatest  common  divisor,  (x  +  1)'.  Hence,  the  given  equation  eon- 
tains  four  values  equal  to  —  1,  and,  by  dividing  the  equation  by  (x  -f- 1)^, 
we  find  the  other  value  to  be  +  5. 

3.  Required  the  equal  values  in  the  equation,  x*  —  2x^  -f  1  =  0. 

Ans.  Two  values  =  +  1  and  two  values  =  —  1. 


GENERAL  THEORY  OF  EQUATIONS.         461 

a;" 

4.  Required  all  the  values  of  the  equation,  x^  —  —  —  2x^  +  a;^  +  x 

—  J  =  0.  Ans.   +  1,  +  1,  +  J,  —  1  and  —  1. 

5.  Required  all  the  values  of  the  equation,  x*  ■\-  x^  —  6x^  —  4x 
4-8  =  0.  Ans.  +  1,  +  2,  —  2,  —  2. 

After  dividing  out  the  equal  factors,  (x  +  2)  (x  +  2),  we  had  a 
quadratic  left,  which  was  solved  by  known  rules. 

6.  Find  the  values  of  the  equation,  x^  —  .r"  —  2x^  +  2x^  +  x  —  1 
=  0.  J[ns.  +  1,  +  1,  +  1,  —  1,  and  —  1. 

The  first  greatest  common  divisor  is  x^  —  x^  —  a:  +  1.  As  this, 
when  placed  equal  to  zero,  is  too  high  to  solve,  we  find  the  greatest 
common  divisor  between  the  second  differential  coefficient  of  the  given 
equation,  and  the  first  member  of  that  equation,  or,  what  amounts  to 
the  same  thing,  the  greatest  common  divisor  between  x^  —  x^  —  x-\- 1, 
and  its  derived  polynomial,  Zx^  —  2a;  —  1.  This  common  divisor  is 
(a;  —  1)^  Hence,  the  given  equation  contains  three  values  equal  to 
plus  one,  and,  by  dividing  out  by  the  factors,  (a;  —  1)  (x  —  1)  (a;  —  1), 
or,  x^  —  Zx^  +  3a;  —  1,  we  have  left,  a;^  +  2a;  +  1  =  0.  Hence,  the 
other  two  values  arc,  —  1  and  —  1. 

7.  Solve  the  equation,  a;^  +  a;^  —  5a;'  —  a;^  +  8a;  —  4=0. 

Ans.  a:  =  +  1,  +  1,  +  1,  —  2,  —  2. 

8.  Find  the  five  values  of  the  equation,  a;^  +  2a;^  —  16a;  —  32  =  0. 

Ans.   +2,-2,-2,  +v/^=4;  —  v/^iT 

9.  Find  the  values  of  the  equation,  x^  -\- x^  —  2a*  —  16a:^  —  16x  + 
32  =  0.  ^7w.   +  1,  +  2, —2,  —  2,  +^"=^47—^:34: 

10.  Find  the  values  of  the  equation,  x"  —  (2a  +  h)x^-\-  (0^+  2ah)z 

—  en  =  0.  Ans.   +  a,  +  or,  +  J. 

11.  Solve  the  equation,  x^  —  2  (a  +  i)x='  +  (a^  +  4ai  +  l'^)x^—.2 
(a'h  +  aW)x  +  aW  =  0.  Ans.  +  a,  -\- a,  +  b,  +  b 

39* 


462  GENERAL  tii:kory  of  equations. 


DERIVED  POLYNOMIALS. 

476.  We  have  liad  occasion  to  use  the  relation, x  =  a-\-  i/,m  trans- 
forming the  general  equation  in  x  into  another  in  i/,  such,  that  the 
second  term  was  wanting.  But  the  transformation  in  terms  of  y  and  a 
is  of  more  general  application.  Sometimes  a  is  an  undetermined  con- 
stant, whose  value  is  afterwards  determined  in  such  a  manner  as  to 
make  the  equation  fulfil  a  given  condition.  Sometimes  a  is  a  deter- 
minate number  or  quantity,  which  expresses  the  difference  between 
the  values  of  the  given  equation  and  another  which  we  wish  to  form. 
Suppose,  for  instance,  we  wished  to  transform  the  equation,  x^  —  x'^  + 
4a;  4- 5  =  0,  into  another  in  ?/,  such,  that  each  value  of  x  should  exceed 
by  2  each  corresponding  value  of  i/.  Then,  a  must  be  made  equal  to 
2  in  the  relation,  x  =  a  +  i/,  and  we  must  substitute  2  +  y  for  x 
wherever  it  occurs.  Hence,  the  equation  in  7/  would  be  (2  -f-  i/y  — 
(2  -f  7/y  +  4(2  -|-  y)  -f  5  =  0,  or,  expanding  and  reducing,  ^  -f-  5^^ 
-f  12y  -f  17  =  0.  In  this  case,  the  transformation  is  easily  effected 
by  actually  substituting  the  value  of  x  in  terms  of  y,  and  developing 
the  several  binomials  in  the  new  equation.  But,  when  the  given  equa- 
tion is  of  a  high  degree,  this  process  is  tedious  and  impracticable. 
Suppose,  for  example,  it  were  required  to  transform  the  equation,  a;'^  — 
40a;"  +  x}°—2x^  +  8a;^  -f-  5x'  —  4a;«  +  9x'  +  12.r^  —  x^-{-Sx  +  15 
=  0,  into  another  in  y,  such,  that  the  values  of  x  should  be  less  by 
unity  than  the  corresponding  values  of  y.  Then,  a  =  —  1  in  the  re- 
lation, X  =  a  +  ^,  and  we  must  substitute  for  x,  wherever  it  occurs, 
(y —  1).  It  is  plain  that  the  development  of  the  several  binomials  in 
the  new  equation  will  be  exceedingly  tedious. 

The  method  of  derived  polynomials  enaUes  us  to  effect  the  required 
transformation  in  y  without  suhstiiuiion  and  development. 

Let  VIS  resume  the  equation, 

^m   ^  pa;n.-l   _|_  Qa;m-2 -f   To!  -f  U  =   0, 

and  let  us  suppose,  as  before,  x  =  a-{- 1/. 

The  new  equation  in  y  will  then  be 
(a  +  2/T  +  P(«  +  ^/r-'  +  Q(«  +  y)"-' T(a-f  y)+  U  =  0. 

Let  us  assume 
(a  -f  2/r  +  P(a  +y)-'  +  Qia  +  ijT-' T(«.  -f  2/)  +  U  =  A 


GENERAL  THEORY  OF  EQUATIONS.         463 

Making  y=0,  we  get 

a"  +  Pa"-'  +  Qa"-' +  Ta  +  U  =  A. 

Differentiating  tlie  given  equation,  and  dividing  by  dy,  we  get 

K«  +3/)"-'  +  P(»^  — 1)  («  +  VT-''  +  Q("^  — 2)  (a  +  yY^ 

-I-  T  =  B  +  2Cy  4-  3D/  +  4E/  (R). 

Now,  make  y  =  0,  and  denote  by  A'  what  B  becomes  in  tbat  case, 
we  get 

ma-"-'  4-  (wi— 1)  Pa'"-^  +  (m— 2)  Qa""-^ +  T  =  B  =  A'. 

Differentiating  (R),  we  get 

m(m  —  1)  (a  +  ^)"-2  +  (m  —  1)  (m  —  2)  P(a  +  y)"-^  4.  (m  —  2) 
(m  —  3)  Q(a  +  y)°'-»  +  &c.  =  2C  +  2  .  3Dy  +  3  .  4E/  +  &c.  (S). 

Making  y  =  0,  and  representing  by  A"  what  2C  becomes,  we  get 

m(m  —  1)  a—2  +  (m  —  1)  (m  —  2)  Pa""^  +  (w  —  2)  (m  —  3)  Qa"-* 
+  &c.  =  A"  =  2C. 

Then  C  ==  —  ^^  (»^— 1)  ft"""  +  (^?^  —  1)  (m  —  2)  Pg-"-^  +  &c. 

2  2 

Differentiating  (S),  we  get 

m(m  —  1)  (m  _  2)  (a  +  y)"-^+  (m  —  1)  (m  —  2)  (m— 8)  P(a+^)°'--» 
+  (m  —  2)  (m  —  3)  (m  —  4)  Q(a  +  y)""^  +  &c.  =  2  .  3  .  D  + 
2  .  3  .  4Ey  +  &c. 
Making  3/  =  0,  and  preserving  a  similar  notation,  we  have 

_  A^'    _  ?7i(m— 1)  (m— 2)  a"-^+(ni— 1)  (m  — 2)  (m  — 3)  Pa^-^ 

2.3  273 

(«i  —  2)  (m  —  3)  (m  —  4)  Qa-' 
H 2~-g h  &c. 

By  proceeding  in  the  same  way,  all  the  other  coefficients  could  be 
determined.  But  the  law  of  formation  is  already  apparent,  A  is  what 
the  given  equation  becomes  when  a  (or  the  difference  between  x  and^) 
is  substituted  for  x ;  A'  is  formed  from  A,  by  multiplying  the  coeffi- 
cient of  a  in  every  term  by  its  exponent,  and  then  diminishing  that 

A" 

exponent  by  unity ;  -^  is  formed  in  precisely  the  same  way,  except 

A'" 
that  the  result  is  divided  by  2 ;    -5-  is  formed  in  like  manner  from  A", 
o 

except  that  the  result  is  divided  by  3. 

A"" 

-J-  is  formed  according  to  the  same  law  from  A'",  the  result  being 

divided  by  4. 


464 


GENERAL  THEORY  OF    EQUATIONS. 


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GENERAL  THEORY  OF  EQUATIONS.         465 

477.  To  transform  an  equation  into  another,  in  which  the  values  of 
the  new  unknown  quantity  shall  differ  from  those  of  the  old  by  some 
constant  quantity,  we  have  then  this 


RULE. 

I.  Replace  x  in  the  given  equation,  wherever  it  occurs,  hy  the  assumed 
difference,  and  call  the  result  A. 

II.  Find  the  derived  polynomial  of  A,  and  call  it  A'. 

III.  Fiiid  the  derived  polynomial  of  A',  and  call  it  A". 

IV.  Find  all  the  other  derived  polynomials  in  the  same,  and  calcu- 
late their  values. 

V.  Substitute  the  values  so  found  in  the  formula,  A  +   A'^  + 

1 — 9  "^  1 — 9^ — R  '^  ^^'  ^^  ^'  ^"^^  '^*^  transformed  equation  will  he 
found. 

EXAMPLES. 

1.  Transform  the  equation,  x^  —  a;^  +  4x  +  5  =  0,  into  another  in 
y,  so  that  the  values  of  x  shall  exceed  those  of  y  by  2. 

Then  we  have,  by  the  rule, 

(2/  —  (2/  +  4  (2)  +  5  =  17  =  A, 
3(2)2  — 2  (2) +4  ^12  =  A,' 

6  (2)  —  2  =  10  =  A", 

6  =    6  =  A'", 

0  =    0  =  A'-. 

Hence,  the  formula,  A  +  A'y  +  — ^  +  - — -^  +  &c.  =  0,  gives 

17  +  12y  +  5y^  +  ?/'  =  0 ;  or,  changing  the  order  of  the  terms,  i^  -\- 
5y  4-  12y  +17  =  0,  the  same  as  before  found. 

2.  Transform  the  equation,  x^  —  5x'  +  4*^  +  x  +  1  =  0,  into  an- 
other in  y,  so  that  the  values  of  x  shall  exceed  those  of  y  by  unity. 

(1)6  _  5  (1)5  _j.  4  (1^.  ^  (1).  ^  1  _  2  =  A, 
6  (1)^  —  25  (1)^  +  8  (1)  +  1  =  — 10  =  A', 
30  (1)*  — 100  (I)''  +  8  =  —  62  =  A", 
120  (1/  —  300  (1)^  =  _  180  =  M", 
360  (Xf  _  600  (1)  =  —  240  =  A'^ 
720  (1)  =  720  =  A^, 
720  =  720  =  K-K 

2e 


466         GENERAL  THEORY  OF  EQUATIONS. 

Hence,  tlie  transformed  equation  is 

2  _  10^  —  31/  —  SOj/^—  lOy  +  Qy'  +  y  =  0, 
or,  /  +  Q,y'— 10/  —  30/  _  31^/^  —  lOj^  +  2  =  0. 

3.  Transform  the  equation,  x* —  8aj'  +  Ox^  +  £c  +  12  =  0,  into  an- 
other, whose  second  term  shall  be  wanting. 

Ans.  /  —  15/  —  21y  +  2  =  0. 
In  this  example,  x  =  y  —  (~|)  =  y  +  2  (Art.  455). 

4.  Transform  the  equation,  2a;'  -\-  x^  -\-  x  —  4  =  0,  into  another, 
whose  second  term  shall  be  wanting.         Ans.  y^  +  ■f'^y —  If  i  =  0- 

First  divide  the  given  equation  by  2,  to  bring  it  under  the  proposed 
form.     Then,  make  x  =  y  — 3  =  y —  ^. 

5.  Transform  the  equation,  —  x^  -\-  x^-^-  x  —  1  =  0,  into  another, 
whose  second  term  shall  be  wanting.  Ans.  /  — 1?/+  J-|  =  0. 

First  bring  the  equation  under  the  proposed  form  by  multiplying  by 
minus  unity.     Then,  make  x  =  y  -\-  ^. 

6.  Transform  the  equation,  x^  —  ISx^  +  135x^  +  x'  —  a;^  +  2x  +  4 
=  0,  into  another,  whose  second  term  shall  be  wanting. 

Ans.  /  +  541/  +  3653/  +  8771^/  +  7318  =  0. 

Why  did  the  second  and  third  terms  disappear  together  ? 

7.  Transform  the  equation,  x'  —  1x^  +  x^  —  2x*  +  x^  —  Sx^  +  x 
-J-  8  =  0,  into  another  in  y,  such,  that  the  values  of  y  shall  be  less  by 
unity  than  those  of  x. 

Ans.  /  —  20/  —  67/  — 102/  —  86/  —  40y  =  0. 

Why  is  the  absolute  term  wanting  in  the  transformed  equation  ? 
Why  did  the  second  term  disappear  ? 

8.  Transform  the  equation,  4x''  —  3a;^  +  2x  —  3=  0,  into  another 
in  y,  such,  that  the  values  of  y  shall  be  greater  by  2  than  those  of  x. 

Ans.  4/  —  27/  -f  Q2y  —  51  =  0. 

It  will  be  seen  that  +  1  is  a  value  of  x  in  the  given  equation,  and 
+  3  is  a  value  of  y  in  the  transformed  equation,  as  it  ought  to  be. 

9.  Transform  the  equation,  Ax*  —  Sx'  +  2x^ — Sx  =  0,  into  another 
in  y,  such,  that  the  values  of  y  sTiall  be  less  by  unity  than  those  of  x. 

Ans.  4/  +  18/  +  17/  +  lOy  =  0. 


GENERAL    THEORY    OF    EQUATIONS.  467 

Why  is  the  absolute  term  wanting  in  tlie  transformed  equation? 
Why  are  all  the  terms  positive  ? 

10.  Transform  the  equation,  5.r^  —  50x^  +  SOOx^  +  9x^  +  a-  — 165 
=  0,  into  another  in  y,  such,  that  the  values  of  y  shall  be  less  by  2 
than  those  of  x.  Ans.  hrf  +  40V  +  1237y  +  838  =  0. 

Why  did  the  second  and  third  terms  disappear  ? 


EQUATION  OF  DIFFERENCES. 

478.  We  have  now  shown  how  all  the  rational  values  could  be  ob- 
tained, either  by  operating  directly  upon  the  given  equation,  when  of 
the  proposed  form,  or  upon  the  transformed  equation,  and  then,  by  means 
of  the  relations  between  the  values  in  the  transformed  and  given  equa- 
tions, ascertain  the  rational  values  in  the  given  equation.  If  we  would 
next  divide  the  given  equation  by  the  factors  corresponding  to  its  ra- 
tional values,  and  place  the  quotient  resulting  equal  to  zero,  we  would 
have  a  new  equation  containing  only  irrational  and  imaginary  values. 
Before  explaining  the  method  of  finding  the  irrational  values,  it  becomes 
necessary  to  show  how  to  find  an  equation,  whose  values  shall  be  equal 
to  the  difierence  of  values  in  the  given  equation.  Such  an  equation  is 
called  the  equation  of  differences. 

Let  x',  x",  x'",  &c.,  be  values  in  the  given  equation,  and  i/  a  value 
in  the  equation  of  difiFerences  sought.  It  is  plain  that  this  equation 
will  be  of  the  same  form,  whatever  pair  of  values  we  assume  in  the 
given  equation.  Hence,  we  may  assume  y  =  oo'  —  x\  then  a;"  = 
y  -f  cc'.     Since  x"  is,  by  hypothesis,  a  value  in  the  given  equation,  .t° 

+  Px""-*  -\-  Qx'"-2 _|-  Ta;  +  U  =  0,  by  substitution,  we  will 

have  a;'"»  -f  Px""-'  -f  Qx""-^ Trc"  +  U  =  0.      Now,  if  we 

replace  aj"  by  y  4-  ^'  in  this  equation,  and  develop  by  the  formula  in 

7,2 

Article  477,  the  new  equation  will  become  Xq  +  X^y  -\-  Xarp— ^  +  X3 

rz — ^ — ^  -j-  &c.  =  0 ;  in  which  Xq  denotes  what  the  given  equation 

becomes  when  x'  is  substituted  for  .r,  X,  is  the  derived  polynomial  of 
Xoj  X2  the  derived  polynomial  of  X„  &c.,  &c.  But,  since  x'  is,  by 
hypothesis,  a  value,  Xo  is  equal  to  zero,  and,  by  dividing  by  y,  the  new 

equation  becomes  X,  +  ^^zr~  -f  X3:j — '^ — -x  -f  &c.  =  0  (D),  and  is 


468         GENERAL  THEORY  OF  EQUATIONS. 

tlie  equation,  which  combined  with  the  given  equation,  will  give  the 
one  sought. 

479.  It  is  obvious  that,  to  obtain  (D),  we  have  only  to  get  the  deri- 
vative of  the  first  member  of  the  given  equation,  add  to  this  the  second 
derived  polynomial,  multiplied  by  the  first  power  of  y,  divided  by  1  .  2, 
plus  the  third  derived  polynomial,  multiplied  by  rf,  divided  by  1 .  2  .  3, 
plus  other  derived  polynomials,  multiplied  by  corresponding  powers  of 
y  with  their  appropriate  divisors. 

As  a  simple  illustration,  let  it  be  required  to  find  the  equation  of 
diiferences  to  the  equation,  x^  —  4  =  0.  Then  X,  =  2x,  X2  =  2,  and 
(D)  becomes  2a3  +  y  ==:;  0.  Eliminating  x  between  these  equations, 
by  the  method  of  the    greatest  common  divisor,  we  have, 

4x2  —  lQ\2x-\-y 

4lx^  -f  2xy     2x  —  y  =  quotient. 

—  2x1/  — 16 

—  2xi/  —  f 

1/^  —  16  =  0,  equation  of  difi"erences. 

The  values  of  y  in  this  equation,  are  +  4  and  —  4,  and  those  of  x 
in  the  given  equation,  +  2  and  —  2,  the  difference  between  which  is 
either  +  4  or  —  4. 

If  in  the  equation,  7/^  — 16  =  0,  we  make  9/^  =  z,  we  will  have  z  — 
16  =  0,  and  the  value  of  z  will  be  equal  to  the  square  of  the  difference 
of  the  values  of  x.  Such  an  equation  is  called  the  equation  of  the 
square  of  the  differences. 

The  degree  of  the  equation  of  differences  will  be  expressed  by  the 
number  of  combinations  of  m  values,  taken  two  and  two,  or  by 
m^m  —  1),  and,  since  this  product  is  always  even  whatever  may  be  the 
value  of  m,  the  sought  equation  will  always  be  of  an  even  degree. 

As  a  second  illustration,  let  it  be  required  to  find  the  equation  of 
differences  to  the  equation,  x^  —  x  —  6  =  0.  Then,  X,  =  2>x'^  —  1, 
X2  =  6x,  X3  =  6,  X4,  X5,  &c.  =  0,  and  equation  (D)  becomes  ox^  —  1 
-\-  Zxy  +  if  =  ^ ;  or,  Zx^  +  Zxy  -\-  if  —  1  =  0 ;  and,  preparing  the 
given  equation  for  division  by  multiplying  by  y,  we  have 


GENERAL    THEORY    OF    EQUATIONS.  469 

3X'  —  3a;  —  18  |  3z'+3g  y+y'— 1 
3x^+3x^!/+xy^—x  a>-]/=Quotient. 
—3x^y—(y''+2ix—18  I  Sx-'+Zxy+y^-l 

-Sx^y-Sxy'^y'+y  2(jf»-l) 

2{y''—\)x  +  y^  —  y—l'i  \  6(y'^—l')x^+(5iy^—l)xg+2{y'*—l)^\  2d  Quotient  =  3x. 
e(y'^—l)x^+3y^x—Sxy—Ux 
3{y''—l)xy+biv-\-2{y^—l)   =  2d  Remainder. 

or,  3(y^—y+lS)x+2(y^r)'^       Multiplying  by  2(y«-l) 

2(1/^1) 

Hy'  -  y  +  18)  x(y»  -  1)  +  4(y»  -  ir        I  2(y'^l)x+y°-y-I8 
^jy'—y + 1 8)ar(ya— 1  )+3[  y'— y + 18)( y'-y— 1 8)  |  3(y=—y+18)=:3d  Quotient. 
4(3/'  — l)'-3(y'-y  +  18)(y»-y  — 18)=0 

or,  developing  4(/  —  3^/*  +  fi/  —  1)  —  ^f  +  St/*  +  5ii/^  +  St/*  — 
3/  — 54j^  — 54y  +  54y'+  3(18)^  =  0  ;  or,  reducing  /  — 6y^  +  V 
4-  968  =  0,  wliicli  is  the  equation  of  differences  sought. 

Now,  make  y^  =  z,  and  we  have  for  the  equation  of  the  square  of 
the  differences  a'  —  6^^  +  9z  +  968  =  0. 

In  the  above  development,  the  alternate  terms  were  struck  out.  This 
was  to  have  been  anticipated  from  what  had  been  said.  But  we  may 
show,  in  another  manner,  that  the  equation  of  differences  can  contain 
only  even  powers  of  y. 

For,  let  a,  b,  c,  d,  &c.,  be  the  numerical  differences  of  the  values  of 
the  given  equation,  then,  +  «,  —  a,  -f  i,  —  h,  &c.,  will  be  values  in  the 
equations  of  differences;  and,  since  the  factors  corresponding  to  these 
values,  permuted  in  pairs,  give  us  (t/  +  a)  (3/  —  a),  (j/  +  l)  (j/ — J),  &c., 
the  first  member  of  the  equation  of  differences  will  be,  (jj^  —  a^  (^'—-^^) 
(/-cO,&c. 

Now,  making  ^  =  z,  we  will  have  (z  —  a")  (~  —  l")  (z  —  c^)  &c.  =  0 
for  the  equation  of  the  square  of  the  differences.  The  degTce  of  the 
last  equation  will  only  be  half  as  great  as  that  of  the  equation  of  dif- 
ferences. 


EXAMPLES. 

1.  Given,  x^  —  9  =  0,  to  find  the  equation  of  differences. 

Ans.  /  —  86  =  0. 

2.  Given,  x'  +  9x  4-  4  =  0,  to  find  the  equation  of  differences. 

Ans.  /  +  54/  +  81/  +  3348  =  0. 

After  the  first  division  the  remainder  will  be,  2(//^  ^   9)  a;  +  y  +  9?/ 
+  12.       Multiply  the  last   divisor  twice   by   2(/  +  9).       Then, 
after  two  divisions,   you   will   have   4(/  +  9)'  —  3(/  +9y  —  12) 
(2/^  +  ^2/  +  12),  which  will  reduce  to  the  expression  above. 
40 


470         GENERAL  THEORY  OP  EQUATIONS. 

3.  Given,  x^+  ax  +  J  =  0,  to  find  the  equation  of  differences. 

Ans.  if  +  Qai/  -\-  Da^  if  +  4a^  +  27b"-  =  0. 
First  remainder  is,  2(y^  +a'^)x  +  ?/  +  oy  +  oh.  Use  2(y^  +  aF)  twice 
as  a  multiplier. 

4.  Given,  x^  —  a;  =  0,  to  find  tte  equation  of  differences. 

Ans.  y^  —  61/  +  dy'  —  4  =  0. 
First  remainder  is,  2(y'^  —  1)  x  +  y^  —  y.     The  coefiicient  of  x  is 
used  twice  as  a  multiplier. 

It  will  be  seen  that  +  1  and  —  1  are  values  in  the  equation  of  dif- 
ferences. Dividing  by  y^  —  1,  and  solving  the  resulting  equation,  y'*  — 
by'^  +  4  =  0,  by  the  rules  for  binomial  equations,  we  get  the  four  values, 
-f  2,  —  2,  +  1  and  —  1.  The  six  values,  then,  in  the  equation  of  dif- 
ferences, are,  +  2,  —  2,  and  +  1  and  —  1,  repeated  twice.  These  ought 
to  be  the  values,  since  those  in  the  given  equation  are,  0  -f-  1  and  —  1, 
the  differences  between  which  are  those  given  above. 

IRRATIONAL  VALUES. 

480.  An  equation,  freed  from  the  factors  corresponding  to  the  ra- 
tional values,  will  contain  only  irrational  or  imaginary  values,  or  both 
irrational  and  imaginary  values.  We  can  best  explain  the  process  of 
finding  the  irrational  values  by  an  example. 

Let  us  take  the  equation,  x^  —  2x'^  —  2  =  0,  to  find  one  positive 
rational  value.  The  superior  positive  limit  is,  1  +  1/2  =  3.  Substi- 
tuting the  natural  numbers  from  0  up  to  3,  we  find  that  0  and  1  give 
negative  results  when  substituted  for  x  in  the  equation,  and  that  2 
gives  a  positive  result.  Hence,  a  value  of  the  equation  lies  between  1 
and  2,  and  1  is  the  entire  part  of  the  irrational  value. 

Now,  make  x  =  y  +  liu  the  given  equation,  then  the  new  equation 
in  y  will  have  values  less  by  unity  than  those  in  x.  Hence,  y  will  con- 
tain the  decimal  part  of  the  value  of  x.  This  transformation  can  be 
most  readily  effected  by  the  formula  of  Article  477.     We  have 

(1)5  — 2(1)^  —  2  =  —3  =  A. 
5(1)*  — 6(1)2  =  — 1  =  A'. 
20(1)3  — 12(1)' =  8  ^j^n^ 
60(1)2  _  12  =  48  ==  A'". 
120(1)»  =  120  =  A-. 
120  =  120  =  A^ 


GENERAL  THEORY  OF  EQUATIONS.         471 


Then,  A  4-  AV  +  1-^  +  ^i-^  + 


1.2    '    1.2.3   '    1.2.3.4   '    1.2.3.4.5 

+  &c.  =  0  becomes,  —  S  —  y  +  Ay''  +  Sf  +  5y*  +  y^  =  0 ',  or, 
changing  the  order  of  terms,  y^  -f-  5y*  +  Sy^  +  4?/*  —  y  —  3  =  0. 

If,  now,  we  make  z  =  lOy,  then  y  =  — .     The  first  figure,  then,  of 

the  value  of  z,  in  the  transformed  equation,  will  be  tenths  in  the  value 
of  y ;  and,  consequently,  tenths  in  the  value  of  x.  The  transformed 
equation  (Article  467)  is, 

^  +  502"  +  800;^^  +  4000;^^  _  lOOOOj/  _  300000  =  0. 
S.  p.  Limit  =  1  +  V300000  =  1  +  23  =  24. 

The  limit  is  here  too  great  for  any  practical  use.  Substituting,  be- 
tween 0  and  10,  we  find  a  change  of  sign  in  the  results  corresponding 
to  the  substitution  of  5  and  6.     For  5,  we  have, 

(5)H50(5)"+800(5)'+4000(5/— 10000(5)  — 300000=— 115625. 

And  6  gives, 

(6)H50  (6)"+ 800  (6)='+4000  (6)^-10000(6)— 300000  =  +  29376. 

Hence,  5  is  the  entire  part  of  the  value  of  z,  and  this  corresponds  to 
five-tenths  in  y  and  x.  Tliereforc,  1  .  5  is  the  approximate  value  of  a;, 
to  within  tenths.  To  get  a  nearer  approximation,  let  us  transform  the 
equation  in  z  into  another  in  w,  so  that  the  values  of  to  shall  be  less  by 
5  than  those  of  z.  Then  iv  will  contain  the  remaining  part  of  the  deci- 
mal value  of  z.     The  equation  in  w  is, 

lo'  -f  75^"  -f  2050w'  +  24750ii;='  -f  118125io  — 115625  =  0. 

Making  t  =  10?p,  or  lo  =  — ,  the  transformed  equation  in  t  (Art.  467), 

will  be, 

i*-f750<"+205000i=' -1-24750000/2+1181250000^— 11562500000  =  0. 

On  trial,  we  find  that  8  and  9  give  results  with  contrary  signs ;  hence, 
8  is  the  entire  part  of  the  value  of  t,  and  corresponds  to  -8  in  w,  and 
to  .08  in  y  and  x.  We  have,  then,  for  the  approximate  value  of  x, 
1-58.  The  process  would  be,  in  all  respects,  the  same,  were  a  negative 
value  to  be  found,  except  that  we  would  find  the  negative  limits. 

Let  us  take,  as  a  second  example,  x^  —  lOx'  -{-  6x  +  1  =  0. 

We  find  that  0  and  1  give  results  with  contrary  signs,  and  so  do  3 


472         GENERAL  THEORY  OF  EQUATIONS. 

and  4.  Let  us  find  the  decimal  part  of  the  value,  whose  entire  part  ia 
3.     Mate  x  =  y  +  S.     The  equation  in  y  will  be, 

rf  +  m/  +  SOy'  +  ISOy"  +  Uly  -8=0, 

and  that  in  z, 

z^  +  I50z*  +  80002»  -f  ISOOOOz^  +  UlOOOOz  — 800000  =  0. 

Substituting  0  and  1,  we  find  a  change  of  sign.  Hence,  0  is  the 
tenths  of  the  given  equation.     Making  z  =  0  +  «;,  we  have, 

to'  +  IbOw*  +  8000!«='  +  180000i«^  +  1410000«;  —  800000  =  0. 

The  equation  in  t,  is, 

/'+1500<''+800000f^4-180000000«^+ 14100000000^— 80000000000=0. 

We  find  that  5  and  6,  when  substituted,  give  a  change  of  sign. 
Now,  make  f  =  s+  5,  the  equation  in  s  will  be, 

.s5_^175s^4-830250s^+192226280s2+15960753125s— 4899059375=0. 

By  changing  this  into  an  equation  in  r,  making  r  =  10s,  we  will  find  a 
change  between  3  and  4.  Hence,  3  053  is  the  approximat-e  value  of  a; 
to  within  thousand tlis. 

481.  In  this  process  we  have  proceeded  upon  the  hypothesis,  that 
but  one  real  root  lay  between  two  successive  integers.  To  ascertain 
whether  this  is  the  case,  we  have  only  to  transform  the  given  equation 
in  X  into  another  in  y,  so  that  the  values  of  y  shall  be  the  squares  of 
the  difi"erences  of  those  of  x.  Next,  find  the  inferior  limit  of  the  posi- 
tive values  of  y.  Suppose  D^  to  be  this  limit ;  then,  since  D^  is  less 
than  the  least  value  in  the  equation  of  the  square  of  the  differences, 
v^D^,  or  D,  will  be  less  than  the  least  difference  between  the  values  in 
the  given  equation.  Now,  if  D  be  ^  1,  it  is  plain  that  no  real  root 
will  be  comprised  between  two  successive  integers,  and  the  process 
described  above  can  be  pursued.  A  similar  course  of  reasoning  can 
be  applied  to  D'^,  the  inferior  limit  of  the  negative  values  in  the  equa- 
tion of  the  squares  of  the  differences. 

But,  if  D  <^1,  then  two  or  more  real  roots  may  be  comprised  be- 
tween two  consecutive  integers.  In  this  case,  we  have  only  to  substi- 
tute a  series  of  numbers,  whose  common  difference  shall  be  =  or  <^  D. 
Then,  those  numbers,  which  give  results  with  contrary  signs,  will  have 
but  one  real  value  between  them.     Another  method  of  frequent  appli- 


GENERAL  THEORY  OF  EQUATIONS.        473 

cation,  when  D  is  a  proper  fraction,  -,  is  to  transform  the  equation  in  x 

into  another  in  y,  by  making  x  =  -.     Then,  the  differences  of  the  values 

of  y  will  be  greater  than  unity,  and  only  one  real  root  will  lie  between 
the  successive  integers  in  the  transformed  equation.     For,  let  x'  and 

x"  be  consecutive  values  of  x,  then  x'  =  -,  and  x"  =  — .     Hence, 

r  r 

(x'  —  x")  r  =  y'  —  y".     Then  the  differences  between  the  consecutive 

values  of  2/  is  r  times  greater  than  between  those  of  x,  and,  as  r  is  the 

denominator  of  D,  y'  —  y"  must  be  greater  than  unity. 


EXAMPLES. 

1.  Find  one  irrational  value  in  the  equation,  x^  —  8.1'+  Ix^  —  56  =  0. 

Ans.  2-828. 

2.  Find  one  irrational  value  in  the  equation,  a;*  +  3ic'  —  4x^  —  15a; 

—  5  =  0.  Ans.  2-236. 

3.  Find  one  irrational  value  in  the  equation,  x^-\-  2x' —  2ar^  —  4  =  0. 

Ans.  1-264. 

4.  Find  one  irrational  value  in  the  equation,  .x'  —  1x  +7  =  0. 

Ans.  —3-048. 

5.  Find  one  irrational  value  in  the  equation,  x*  -\-  x^  —  12:c^  —  17a; 

—  85  =  0.  Ans.  4-123. 

6.  Find  one  irrational  value  in  the  equation,  x^  —  2  =  0. 

Ans.  1-414. 

7.  Find,  by  the  process  of  irrational  values,  one  value  in  the  equa- 
tion, x^ —  4  =  0. 

Ans.  x^2,  the  decimal  part  in  all  the  transformed  equations  being 
zero. 

8.  Find  one  irrational  value  in  the  equation,  x*  +  x^ —  25x^  —  26a; 

—  26  =  0.  Ans.  5-099. 

482.  If  we  apply  the  foregoing  process  to  an  example  of  the  form, 

x*  +  14x^  —  49  =  0,  the  consecutive  numbers  will  give  no  change  of 

sign.     One  value  in  the  equation  of  differences  will  be  found  to  be 

zero,  and  also  one  in  the  equation   of  the  square  of  the  differences. 

40* 


474        GENERAL  THEORY  OF  EQUATIONS. 

D'^  and  D  are  tliea  botli  zero.  "When  this  is  the  case,  we  may  infer  the 
presence  of  equal  values.  Ou  trial,  as  in  Art.  475,  we  find  x^  —  7  the 
greatest  common  divisor  between  the  first  member  of  the  given  equa- 
tion and  its  derived  polynomial.  We  see,  by  this  example,  that  the 
preceding  method  is  only  applicable  to  equations  which  have  been  freed 
from  their  equal  values. 

While  the  foregoing  method  afi'ords  a  complete  solution  to  the  pro- 
blem of  finding  the  irrational  values  of  numerical  equations,  yet  it  is 
of  diificult  application  to  equations  of  high  degrees,  and,  in  all  equa- 
tions, whether  of  low  or  high  degrees,  the  difl&culty  increases  with  the 
number  of  decimal  places  sought. 

NEWTON'S  METHOD  OF  APPROXIMATION. 

483.  This  is  known  as  the  method  of  successive  substitutions,  and 
consists  in  substituting,  in  the  given  equation,  the  natural  numbers  be- 
tween the  limits,  until  results  with  contrary  signs  are  obtained.  Let  a 
be  the  least  of  two  consecutive  numbers  which  give  results  with  con- 
trary signs.  Then  a  is  the  first  figure,  or  entire  part  of  the  value 
sought.  Substitute  a  -{-  y  for  x  in  the  given  equation,  y  being  a  small 
fraction,  whose  second  and  higher  powers  may  be  neglected.  Hence,  y 
may  be  found  in  the  transformed  equation,  and  a  +  y  will  constitute 
the  first  approximation  to  the  value  of  x.  Let  a  -1-  y  be  denoted  by  h, 
and  make  x  =  h  -{■  y',  y'  being  a  small  fraction,  whose  higher  powers 
may  be  neglected.  The  transformed  equation  will  give  the  value  of 
y,  and  this,  with  6,  will  constitute  the  second  approximation  to  x. 
Calling  h  +  y',  c,  and  making  x  =  e  +  y" ,  we  can  get  a  third  approxi- 
mation to  the  value  of  x,  and  may  thus  carry  the  approximation  to  as 
many  places  of  decimals  as  we  please. 

Let  us  take,  as  an  illustration,  the  equation,  a;*  —  2  =0.  Substi- 
tuting between  the  limits,  we  find  that  1  and  2  give  different  signs. 
Then  1  is  the  entire  part  of  the  value.  Make  a;  =  1  -f  y,  and  reject 
y^,  we  will  have  1  -j-  2^/  —  2  =  0,  or  y  =  J  =  -5.  Hence,  for  first 
approximation,  x  =  1-5.     Now,  make  x  =  1-5  +y,  and  we  will  have, 

•25 

after  rejecting  y'\  2-25  +  3/  —  2  =  0,  or,  /  =  —  -^  =  —  -0833. 

Then,  for  the  second  approximation,  we  get  cc  =  1  -5  —  -0833  =  1-4167. 
Place  X  =  1-4167  +  y",  we  get 

2-8834/'  =  —  -00303889,  or,  y"  =  —  -00107. 
Then,  for  a  third  approximation,  x  =  1-4167  —  -00107  =  1-41563. 


GENERAL  THEORY  OF  EQUATIONS.         475 

484.  The  accuracy  of  the  process  evidently  depends  upon  the  un- 
known quantity  introduced  being  a  small  fraction.  "Whenever,  then, 
the  substitution  of  the  natural  numbers  make  y  an  improper  fraction, 
more  minute  substitutions  must  be  made,  unless  we  intend  to  carry  the 
approximation  to  several  places  of  decimals. 

Thus,  take  the  example,  x''  +  x  —  8  =;  0,  we  find  that  1  and  2  give 
results  with  contrary  signs.  Then,  1  is  the  entire  part  of  the  value 
sought.  Making  x  =  y  +  1,  we  have,  3/^  +  3y  +  3^^  +  1  -|-  y  +  1 
—  8  =  0,  or,  rejecting  the  higher  powers  of  y,  4^/ —  6  =  0.  Hence, 
y  ■=  &=.  1*5,  an  improper  fraction.  Then,  for  the  first  approximation, 
we  have,  x  ^  2-5,  a  result  obviously  too  great. 

Next,  place  x  =  2-50  +  ?/,  we  have,  after  rejecting  higher  powers 
of  y,  (2-50)»  +  3  (2-50)2y  +  2-50  +  y  —  8  =  0.  Then,  y'  =  —  -512, 
and,  for  the  second  approximation,  x  =  1-988.  Making  now  x  ^ 
1-988  +  y" ,  we  will  have,  after  rejecting  the  higher  powers  of  if' ■> 

(1-988)''  +  3  (l-988)y'  4-  1-988  +  y"—  8  =  0, 
or,  7-856862272  +  ll-856432y'+  1-988  -f  y"  — 8  =  0, 
or,  11-856432/'  =  _  1-844862272. 

Hence,  y"  =  —  -144  nearly,  and,  for  a  third  approximation,  x  =  1-844, 
which  difi'ers  from  the  true  value  by  less  than  one  hundredth.  We 
see,  from  this  example,  that  the  error  was  considerable  when  y  was  an 
improper  fraction,  but  became  reduced  by  carrying  the  approximation 
farther.  A  closer  approximation  could  have  been  obtained,  without 
carrying  the  operation  so  far,  by  making  minute  substitutions.  A 
change  of  sign  would  have  been  found  between  the  results  corresponding 
to  1-8  and  1-9. 

LAGRANGE'S  METHOD. 

485.  This  differs  from  Newton's,  in  that  the  unknown  expression 
added  to  complete  each  successive  value  is  fractional  in  form,  and  in 
that  the  transformation  is  made  into  an  equation  involving  this  un- 
known quantity.     Thus,  let  a  be  the  entire  part  or  number  next  below 

the  value,  we  make  x  =  a  H — ,  and  get  a  transformed  equation  in  y, 

such,  that  the  values  of  y  must  be  greater  than  unity.  Let  h  be  the 
entire  part  of  the  value  of  y,  then,  for  the  first  approximation,  we  have 

a:  =  a  4-  -  =  — .     Next,  place  y  =  &  A — .       The  transformed 

equation  iu  ^  must  have  its  values  greater  than  unity.     Let  c  be  the 


476         GENERAL  THEORY  OP  EQUATIONS. 

entire  part  of  tlie  value  of  y' .     Then,  approximatively,  y  =  h  ■{ —  = 

,  and,  for  a  second  approximation  to  x,  we  have,  x  ^  a  -^^  -  =  a 

c  .  .  1 

+ -.     To  find  a  third  approximate  value  for  x,  let  y  =  c  -| — j^ 

CO  -\-  I  y  1 

and  we  may  thus  continue  to  approximate  to  the  value  of  x  until  the 

result  is  as  accurate  as  desired. 

To  apply  these  principles,  let  us  take  the  equation,  x^  —  x  —  5  =  0, 

We  see  that  1  and  2  give  contrary  signs,  hence,  cc  =  1  +  -.  Substi- 
tuting this  value  of  cc,  we  have  5^^  —  2f/^  —  3y  — 1  =  0.  The  entire 
part  of  the  value  of  y  is  1,  hence,  for  first  approximation  to  x,  we  have 

X  —  \  ■\ —  =  1  +  1  =  2,  which  is  plainly  too  great.     Next,  making 

y  =  1  +  -,  we  get  y'^  —  %y'^  —  13/  —  5  =  0.     In  this,  9  and  10 

give  contrary  signs.  Then,  y  =  1  +  -^  =  Lo,  and  a;  —  1  +  y^^  =  12^ 
fbr  the  second  approximation. 

Now,  make  ?/'  =  9  -| — ^,  and  the  transformed  equation  in  y"  will  be, 

4iy"'  —  86^"^  —  19y'  —  1  =:  0  ;  a  change  of  sign  between  the  re- 
sults given  by  2  and  3.  Hence,  y'  =  9|  =  i^^,  y  =  1  +  -2^  =  |^, 
and  a;  =  1  -f  ^a  =  |fl,  for  the  third  approximate  value  of  x.  We  find 
on  trial  the  third  approximate  value  a  little  too  great,  and  the  second  a 
little  too  small.  Hence,  the  true  value  lies  between  4f  ^^^^  io>  ^^^ 
either  of  these  numbers  diflPers  from  the  true  value  by  less  than  ^  jq. 

486.  Let  us  take  the  simple  example,  x^  —  2  =  0.  Place  x=^\  •\ — , 

y 

then,  y  —  1y  —  1  =  0.     We  find  that  2  and  8  give  contrary  signs. 

Hence,  for  first  approximation,  x  =  1  +  i  =  |.  Now,  make  ?/  =  2  -f-  -» 

and  the  equation  in  y  will  become  y'"^  —  2y  — 1  =  0;  and,  again,  2  is 
the  entire  part  of  the  value.  Hence,  y  =  2  +  i  ==  55  and  x=^\  -\-  \ 
=  I,  for  a  second  approximation.  On  trial,  we  find  |  too  great,  and  | 
too  small,  and  these  differ  from  each  other  by  a  tenth  j  therefore,  the  true 
value  differs  from  either  by  less  than  a  tenth.  For  a  third  approximate 
value  we  would  find,  x  =  |-|,  which  is  a  little  too  great,  but  is  within 
g'jj  of  the  true  value.  The  fourth  approximate  value,  ||,  is  too  small, 
but  differs  from  the  true  value  by  less  than  g-J-,  •  By  continuing  thus 
the  process,  we  would  find  the  approximations  alternately  too  great  and 


GENERAL    THEORY    OF    EQUATIONS.  477 

too  small,  and  thus  can  tell  at  every  step  how  near  we  have  come  to  the 
true  values.  We  see  that  Lagrange's  method  enables  us  to  determine 
the  proximity  to  the  true  value  at  every  stage  of  the  work,  and  this  is 
the  chief  advantage  claimed  for  it  over  the  process  of  Newton. 


EXAMPLES. 

1.  x^  —  7x  +  7  =  0.      Ans.  X  =  f  I,  after  three  approximations. 

2.  X*  —  a-  —  6  =  0.  Ans.  .r  =  |,  after  third  approximation. 
1st  approximation,  2 ;  2d  approximation,  ^  ;  3d  approximation,  |. 

GENERAL   SOLUTION  OF  NUMERICAL   EQUATIONS, 

487.  When  we  have  a  numerical  equation  of  any  degree  to  solve,  we 
must  first  find  its  rational  values  by  the  process  of  divisors,  if  it  is  under 
the  proposed  form.  But,  if  it  is  not,  we  must  transform  it  into  another 
equation  in  y,  so  that  the  coefficient  of  the  first  term  shall  be  plus 
unity,  and  the  other  coefficients  entire.  Then,  knowing  the  relation  be- 
tween 1/  and  X,  we  can  determine  the  values  of  x  when  those  of  y  have 
been  found.  Next  we  must  ascertain  whether  any  of  the  values  are 
repeated  (Art.  475).  Having  thus  found  all  the  rational  values,  we 
next  divide  by  the  factors  corresponding  to  them.  Then,  by  means  of 
the  equation  of  the  squares  of  the  difi'crenccs,  we  can  ascertain  what 
series  of  numbers  to  substitute  in  the  reduced  equation  (Art.  481).  The 
final  step  is  to  find,  approximatively,  the  irrational  values  by  either  of 
the  three  processes  explained.  Now,  if  the  number  of  rational  and  ir- 
rational values  be  subtracted  from  the  degree  of  the  given  equation,  the 
difference  will  be  the  number  of  imaginary  values. 

GENERAL  EXAMPLES. 

1.  Find  the  three  values  of  the  equation,  2x' ~ f-  106a;  — 

20  =  0.  A71S.  a;  =  1,  5,  and  10. 

2.  Solve  the  equation,  x^  —  Ux''  +  7x^  +  28a;  —  28  =  0. 

Ans.  x  =  ±2,  1-356,  1-692,  and  —3-044. 

17a^ 

3.  Solve  the  equation,  x''  -\ — |a; — i  =  0. 


Ans.  X  =  ^,  —  I,  and  —  3. 


478  Sturm's  theorem. 

4.  Solve  tte  equation,  x"  —  lO^c^  —  12^2  +  92x  +  280  =  0. 

A71S.  X  =  10,  and  4-302. 

5.  Solve  the  equation,  2x^  —  2x* —  +  -^  -{■  x  — 1  =  0. 

A71S.  a;  =  1,  ^,  —  J,  and  =h  ,y2. 

6.  Solve  the  equation,  x''  —  4x^  —  4x^  +  Sdx*  —  60x'  —  28a:^  + 
224a;  —  224  =  0. 

7.  Solve  the  equation,  x*  —  2x^  —  ^\x^  +  22a;  +  280  =  0. 

Ans.  X  =  4,  and  5-134. 


STUEM'S    THEOREM. 

488.  When  we  have,  by  means  of  the  process  of  divisors  and  the 
method  of  equal  values,  detected  all  the  rational  values,  and  then,  by 
the  aid  of  the  equation  of  differences,  discovered  all  the  irrational 
values,  we  can  determine  the  number  of  imaginary  values,  by  subtract- 
ing the  number  of  rational  and  irrational  values  from  the  degree  of  the 
equation.  When  this  difference  is  zero,  there  are,  of  course,  no  imagi- 
nary values. 

But  this  method  is  circuitous,  and  is,  moreover,  rendered  tedious  by 
the  employment  of  the  equation  of  differences.  Sturm's  Theorem  de- 
termines directly  the  number  of  imaginary  values,  and  dispenses  tcith 
the  equation  of  differences. 

Let  X  =  x™  -f  Pa;-"-'  -f  Q^c--^  + +Ta:  +  U  =  0  ie  aji  equa- 
tion cleared  of  its  equal  values.     Take  the  derivative  of  X,  and  call  it 

X' ;  then,  X'  =  mx^-'-\-  (m  —  l)Px"»-2-f  (m  —  2)Qx'°-^  -f- -f  T. 

Now,  divide  X  by  X',  and  continue  the  division  until  we  get  a  remaiur 
der  in  a;  of  a  lower  degree  than  X'.  Call  this  remainder,  with  its  sign 
changed,  X".  Divide  X',  in  like  manner,  by  X",  and  continue  the 
division  until  the  remainder  is  a  degree  lower  than  the  divisor.  Change, 
in  the  same  way,  the  signs  of  all  the  terms  of  this  remainder,  and  call 
the  resulting  expression  X'".  Pursue  the  same  process  until  we  get  a 
remainder  independent  of  x,  which  must  be  so  eventually,  since  the 
given  equation,  by  hypothesis,  contains  no  equal  values.  This  will  be 
the  (m  —  1)*''  remainder,  and  all  the  expressions,  including  X  and  X', 


Sturm's  theorem.  479 

will  make  up  (m  +  1)  functions.  Hence,  designate  by  X^+'j  tlie  last 
remainder,  with  its  sign  changed.  We  will  then  have  the  following 
expressions,  Q,  Q',  &c.,  designating  quotients : 

X     =X'Q     -X"  (A), 

X'    =X"Q'    —X'"  (B), 

X"  =  X"'Q"  —  X-  (C), 

X'"  =  X'vQ"'  —  X-  (D), 

X-  =  X-Q-— X-  (E), 


X»-'  =  X-'Q"->— X»+'    (W). 

In  preparing  the  dividends  for  division,  we  may,  as  in  the  process 
for  finding  the  greatest  common  divisor,  multiply  by  any  positive  con- 
stant. In  like  manner,  we  may  suppress  any  positive  constant  common 
to  all  the  terms  of  any  of  the  successive  remainders. 

489.  The  theorem  we  have  to  demonstrate  may  be  enunciated  as 
follows : 

If  we  substitute  any  number,  tn,  for  x  in  the  above  series  of  func- 
tions, X,  X',  X",  &c.,  and  count  the  variations  of  signs  in  the  results, 
and  then  substitute  any  other  number,  as  n,  for  x  in  the  same  series, 
and  count  again  the  variations  of  signs  in  the  results,  the  difference 
between  the  number  of  variations  in  the  tico  cases  icill  exjjress  exactly 
the  number  of  real  values  between  the  numbers  m  a7id  n.  If,  then,  m  be 
taken  as  the  superior  limit  of  the  positive  values,  and  n  as  the  superior 
limit  of  the  negative  values,  the  difibrencc  between  the  number  of  vari- 
ations will  express  the  total  number  of  real  values.  When  this  difiFe- 
rence  is  zero,  there  are  no  real  values ;  when  it  is  equal  to  the  degree 
of  the  equation,  all  the  values  are  real. 

490.  The  demonstration  of  the  theorem  depends  upon  the  four  fol- 
lowing principles : 

1.  M)  tico  consecutive  finctions  can  vanish  for  the  same  value  of  x. 

For,  if  two  functions,  as  X"  and  X'",  could  disappear  together,  equa- 
tion (C)  would  give  0  =  0  —  X'^.  Hence,  X'^  =  0,  and  X"'  and  X*'' 
being  zero,  equation  (D)  would  give  X^  =  0.  And  so  all  the  func- 
tions in  succession  would  become  zero,  until  we  would  finally  have 
X^^'  ^  0.  But  this  would  indicate  a  common  divisor,  and,  conse- 
quently, equal  values,  which  is  contrary  to  the  hypothesis  with  which 
we  set  out. 


480  Sturm's  theorem. 

2,  When  such  a  value  is  given  to  x  as  to  make  any  function  other 
than'K  equal  to  zero,  the  functions  adjacent  to  the  one  disajij^earing  will 
he  affected  with  contrary/  signs. 

For  example,  let  X'^  =  0.  Then,  equation  (D)  will  give  X'"=— X^; 
and,  since  the  own  sign  of  X'^  may  be  either  plus  or  minus,  this  equa- 
tion may  be  written  X'"  =  —  (±  X^).  Now,  if  X'^  be  affected  with 
the  positive  sign,  the  second  member  will  be  negative;  and,  since  the 
signs  of  the  two  members  of  an  equation  must  ^ways  be  the  same,  X"' 
will  be  negative,  and,  therefore,  of  a  contrary  sign  to  X'^.  But,  if  the 
own  sign  of  X^  be  negative,  then  the  second  member  will  be  positive. 
Hence,  X'"  will  be  positive,  and,  therefore,  of  a  contrary  sign  to  X^. 

3.  The  disappearance  of  any  function  other  than  'Kfor  a  piarticidar 
value  ofx,  will  neither  increase  nor  decrease  the  variations  of  signs  in 
the  results  of  the  sej'ies,  X,  X',  X",  d;c. 

Suppose  that  one  of  the  intermediate  functions,  as  X'"',  becomes  zero, 
for  the  value  x  =  b.  Then,  by  the  first  principle,  neither  X'"  nor 
X'^  can  be  zero  for  x  =  I ;  and,  by  the  second  principle,  they  must 
have  contrary  signs.  There  is,  then,  one  variation  of  sign  between  X'" 
and  X"^  when  x  =  b.  We  are  to  show  that  this  variation  was  not 
caused  by  the  disappearance  of  X'''.  For,  however  little  x  ■=h  may 
differ  from  a  value  of  X'"  or  X''^,  h  may  be  taken  smaller  than  this 
difference.  And,  moreover,  since  a  quantity  can  only  change  its  sign 
in  passing  through  zero  or  infinity,  it  is  plain  that,  between  x  =  b  and 
x  =  b  —  h,  neither  X'"  nor  X^  can  undergo  any  change  of  sign.  They 
will  still  be  affected  with  contrary  signs,  and,  for  cc  =  &  —  h,  we  will 
have  either  +  X'",  ±  X'',  —  X^,  or  else,  —  X'",  ±  X'^,  +  X\  We 
affect  X''  with  the  ambiguous  sign,  since  we  do  not  know  whether  its 
sign  is  positive  or  negative.  Now,  read  the  double  sign  either  way, 
and  take  either  of  the  preceding  expressions,  and  you  will  have  one 
variation.  Hence,  before  x  arrives  at  the  value  b,  which  causes  the 
disappearance  of  X"',  the  three  successive  functions  give  one  variation. 
Now,  when  x  passes  beyond  b,  and  becomes  b  +  h  (h  being  less  than 
the  difference  between  b  and  a  value  of  either  X'",  or  X"),  X'"  will 
change  its  sign,  but  X'"  and  X"  will  not.  Hence,  forx:=h-\-  h,  we 
will  have  either  +  X'",  =f  X'",  —  X^  or  —  X'",  =p  X'",  +  X'.  And, 
reading  the  ambiguous  sign  either  way,  in  either  of  the  above  expres- 
sions, we  will  have  one  variation.  Hence,  both  before  and  after  x 
reached  the  value  b,  which  made  X'''  =  0,  we  find  a  variation  of  sign 
among  the  three  functions.  Then  x:=b,  or  the  vanishing  of  the  inter- 
mediate function,  X'\  has  not  caused  that  variation. 


Sturm's  theorem.  481 

4.  Every  passage  ofS.  through  zero  causes  a  loss  of  one  variation 
xchen  x  is  ascending  towards  the  superior  positive  limit,  or  again  of 
one  variation  throughout  the  series,  X,  X',  &c.,  when  x  is  descending 
toioards  the  superior  negative  limit. 

For,  let  a;  =  a  be  a  value  in  the  equation,  X  =  0,  and  take  u  lesp 
than  the  difference  between  a  and  a  value  of  X'.  Then,  X  will  have 
a  different  sign,  when  a  +  u  is  substituted  for  x,  from  what  it  would 
have  were  a  —  u  substituted  for  x.  Hence,  in  the  passage  of  x  from 
a  —  u  to  a  +  M,  X  will  undergo  a  change  of  sign,  but  X'  will  not. 
since,  by  hypothesis,  no  value  of  X'  is  passed  over.  Neither  will  any 
intermediate  function,  X",  X'",  &c.,  undergo  a  change  of  sign  unless  it 
pass  through  zero,  and,  by  the  third  principle,  such  passage  will  not 
affect  the  total  number  of  variations.  Therefore,  in  the  passage  of  x 
from  a  —  «  to  a  +  u,  either  a  permanence,  existing  between  X  and 
X',  is  changed  into  a  variation,  or  else  a  variation  is  changed  into  a 
permanence. 

We  will  now  examine  the  order  in  which  this  change  takes  place. 

Designate  by  X,  what  X  becomes  when  x  =  a  —  u.     Next,  develop 

all   the   terms   of  X   by  Art.  456;    we  will  have  Xi=  A  —  A'm  + 

A"«^         A"'m' 

-j — -y  —  ^j — -^ — r,  +  &c>    (R))  in  which  A  denotes  what  X  becomes 

when  xz=a;  A'  is  the  derivative  of  A,  A"  the  derivative  of  A',  &c. 
But,  since  a  is  a  value  of  .r,  A  must  be  equal  to  zero.     Then,  equation 

(R)  may  be  written,  X,=  —  u  {M  —  i-^  +        ,^"     —  &c.)      Now. 

u  may  be  taken  so  small  that  the  first  term  will  be  greater  than  the 
algebraic  sum  of  all  the  other  terms  within  the  brackets.  Hence,  the 
sign  of  the  quantities  within  the  parenthesis  will  depend  upon  that  of 
A'.  If,  therefore,  the  own  sign  of  A'  is  positive,  the  second  member 
will  be  negative.  Then,  X,  must  be  negative,  and,  therefore,  of  a  con- 
trary sign  to  A'.  But,  if  the  own  sign  of  A'  be  negative,  the  second 
member  will  be  positive.  Then,  X,  will  be  positive,  and  X,  and  A' 
will  be  affected  with  contrary  signs.  And,  since  X,  and  A'  represent 
what  X  and  X'  become  when  a  —  tt  is  substituted  for  x,  we  see  that 
there  is  a  variation  of  signs  between  these  functions  below  a  value, 
x  =  u,  which  satisfies  the  equation,  X  =  0.  Now,  suppose  x  =  a  -\- u, 
and,  therefore,  above  a  value  in  the  equation,  X  =  0.  Develop  as 
before,  designating  by  Xa  what  X  becomes  when  a  -f  u  takes  the  place 

A"?f         A"'u^ 
of  X.     Then,  X2  =  u  (A'  +  ~ — -^  +       ,^    ,^  +  &c.),   in  which  u  is 

41  *2p         " 


482  Sturm's  theorem. 

so  small  that  the  sign  of  A'  controls  that  of  the  parenthesis.  We  see, 
that,  whether  the  own  sign  of  A'  be  positive  or  negative,  Xg  must  be 
affected  with  the  same  sign.  Hence,  when  x  =  a  -{-  ^l,  and,  therefore, 
above  a  value,  there  is  a  permanence  of  sign  between  X  and  X'. 
Therefore,  in  the  passage  of  x  towards  the  superior  positive  limit,  a 
variation  is  lost,  and  changed  into  a  permanence  between  X  and  X', 
whenever  X  becomes  equal  to  zero ;  and,  as  the  other  functions  neither 
gain  nor  lose  variations,  one  variation  only  is  lost  throughout  the  entire 
series  whenever  X  =  0. 

If  we  had  begun  with  a  +  m,  and  passed  down  to  a  —  u,  it  is  evi- 
dent that  one  variation  would  have  been  gained,  every  time  that  x 
reached  a  value  in  the  equation,  X  =  0. 

So,  then,  whenever  x,  increasing  or  decreasing  by  insensible  degrees, 
reaches  a  value  which  will  satisfy  the  equation,  X  =  0,  one  variation 
is  lost  or  gained  in  the  signs  of  the  series,  X,  X',  X",  &c.  Hence,  if 
we  take  any  number,  as  m,  and  substitute  it  for  x  in  all  the  functions, 
and  count  the  number  of  variations  in  the  signs  of  the  results,  and  then 
take  any  other  number,  as  n,  and  treat  it  in  like  manner,  the  difference 
between  the  number  of  variations  in  the  two  cases  will  express  the 
number  of  real  values  lying  in  the  given  equation  between  m  and  7i. 
Moreover,  it  is  plain  that  if  «i  =  +  oo,  and  n  —  —  go,  this  difference 
will  express  the  total  number  of  real  values  in  the  equation.  This 
would  be  equally  true  if  we  used  the  superior  positive  limit  and  the 
superior  negative  limit.  But,  there  is  this  advantage  in  the  employ- 
ment of  -I-  oo,  and  —  oo ;  the  substitution  need  only  be  made  in  the 
leading  term  of  the  successive  functions,  for  then  the  sign  of  this 
term  would  control  the  signs  of  all  the  other  terms  in  the  same 
function. 

491.  Having  determined  the  number  of  real  values,  the  next  step  is 
to  determine  the  initial  figure  of  each  one  of  theSe  values.  To  do  this, 
we  substitute  the  natural  numbers,  0,  1,  2,  3,  &c.,  until  we  get  the 
same  number  of  variations  of  signs  in  the  entire  series  as  was  given  by 
+  00.  The  number  that  gives  the  same  variations,  as  +  oo,  is  the 
superior  positive  limit.      Substitute,   in   like  manner,  0,  — 1,  — 2, 

—  3,  &c.,  until  we   get  the  same  number  of  variations  as  given   by 

—  oo.  We  then  have  the  superior  negative  limit.  The  least  of  the 
numbers,  whether  positive  or  negative,  between  which  a  variation  is 
lost  or  gained,  is  the  initial  figure,  or  entire  part  of  a  real  value. 
Should  there  be  two  or  more  variations  lost  or  gained  between  consecu- 


Sturm's   theorem.  483 

numbers,  then  there  are  as  many  values  as  there  are  changes  of  sign, 
which  have  the  same  initial  figure. 

We  will  illustrate  by  a  simple  example,  -■^'i^ 

Take  equation,  x^  +  3*''  —  1  =  0  =*a:."~  •  Then  its  derivative, 
X'  =  3x2  ^  g^ 

Now,  multiply  all  the  terms  of  X  by  9,  and  divide  by  X'.  After 
two  divisions,  we  will  get  a  remainder  of  a  lower  degree  than  X' ;  this 
is  —  2x  —  9.  Change  its  sign,  and  call  it  X".  Then,  X"  =  2a;  +  9. 
Multiply  X'  by  4,  and,  after  two  divisions,  by  X",  we  will  have  a  re- 
mainder, +  207.     Hence,  X'"  =  —  207,  and  we  will  have 

X     =    x'-f  a;2  — 1=0 
X'   =3^2  + 2a; 
X"  =  2a;  -f  9 
X"  =  —  207. 

Making  a;  =  +  oo,  in  the  leading  terms  of  these  functions,  the  order 
of  the  signs  will  be 

+  +  -1 ,  one  variation. 

And  for  03  =  —  oo,  it  will  be 

1 ,  two  variations. 

Hence,  2  —  1  =  1,  real  value. 

To  find  the  initial  figure  of  this  value,  make  a;  =  0,  1,  2,  3,  &c., 
and  write  the  signs  of  the  results  beneath  their  respective  functions. 
We  will  have 

X,        X',        X",         X'", 
then,    a;  =  0,    —  1,       0,         -f  9,     —  207,  two  variations. 
X  =  1,  ''^•  1,     +5,      +11,    —  207,  one  variation. 

Hence,  1  is  the  superior  limit  of  the  value  sought,  and  0  is  its  initial 
figure.  The  decimal  part  of  the  value  can  be  found  by  the  process  for 
the  irrational  values. 

The  transformed  equation  in  y,\s,  y^  -\-  y^  —  1  =:  0 ;  and  that  in  z, 
z'  +  \^z^  — 1000  =  0.  And,  since  7  and  8  give  results  with  con- 
trary signs,  7  is  the  tenths  of  the  required  value.  Then,  a;  =  0-7  is 
an  approximate  value  of  x. 

The  approximation  may  be  carried  as  far  as  desired  by  the  method 
of  irrational  values. 


484  Sturm's  theorem. 

We  need  not  substitute  0,  —  1,  —  2,  &c.,  in  the  equation,  since 
there  is  but  one  real  value,  and  that  has  been  shown  to  be  positive. 

492.  Whenever  only  one  variation  is  lost  between  two  consecutive 
numbers,  Sturm's  theorem  affords  no  advantage  over  the  process  for 
finding  the  irrational  values,  except  that  of  detecting  directly  the  num- 
ber of  imaginary  values.  But,  when  there  are  two  or  more  real  values 
comprised  between  two  consecutive  numbers,  we  are  enabled  by  means 
of  the  theorem  to  dispense  with  the  equation  of  differences,  which 
otherwise  must  be  employed  to  find  the  decimal  part  of  those  values. 
We  will  now  explain  the  chief  advantage  of  the  theorem. 

Let  us  take  the  equation,  x"  —  loi?  —  1x^  -f  4  =  0.  We  will  have 
the  following  series  of  functions  : 

X  =  cc^  —  2x'  —  2a;='  +  4  =  0, 

X'^Sx"  — 6x2  — 4x, 

X"  =  4x'  +  6x2  — 20, 

X'"  =  —  42x2  —  168x  +  300, 

X''  =  —  2880x  +  3840, 

X'  =  — ,  a  constant. 

+   oo  gives  4-  +  +  H ,  one  variation, 

—  cx)  gives 1 1-  H )  four  variations. 

Hence,  there  are  three  real  values. 
Proceeding  as  before,  we  have 

X,  X',  X",  X'",  X''',  X\ 


When,         X  =  0 
"  x  =  1 

"  x=  2 


+    0     -      0       + 

+ +       + 

+    +     +      +       - 


three  variations, 
three"  variations, 
one  variation. 


Since  +  2  gives  the  same  number  of  variations  as  -f  oc,  it  is  the 
superior  positive  limit.  Moreover,  as  there  are  two  variations  lost  be- 
tween 1  and  2,  there  are  two  real  values  between  them.  Had  we  sub- 
stituted the  natural  numbers,  0,  1,  2,  3,  &c.,  in  the  given  equation,  we 
would  have  found  no  change  of  sign,  because  an  even  number  of  values 
lay  between  consecutive  numbers.  To  detect  these  values,  without 
the  aid  of  Sturm's  theorem,  we  must  either  have  recourse  to  the  equa- 
tion of  differences,  or  to  minute  and  tedious  substitutions. 


Sturm's  theorew.  485 

So,  then,  it  is  evident  that  whenever  an  equation  contains  two,  four, 
six,  or  any  number  of  even  values  between  consecutive  numbers,  Sturm's 
theorem  enables  us  to  detect  these  values  in  the  shortest  possible  manner. 

493.  But,  in  addition  to  this,  the  theorem  gives  us  the  means  of 
finding  the  decimal  part  of  the  irrational  values  in  a  shorter  and  better 
manner  than  by  the  usual  process,  as  we  will  now  show. 

Since  1  is  the  entire  part  of  both  the  irrational  values,  the  first  step, 
according  to  the  usual  process  for  finding  the  irrational  values,  is  to 
transform  the  equation  in  x  into  another  in  y,  so  that  the  values  of  i/ 
shall  be  less  by  unity  than  those  of  x. 

The  transformed  equation  in  y  is 

/  +  5/ +  8/ +  2y ^  -  5y  +  1  =  0, 

and  that  in  z,  is 

z^  +  502"  +  800z'  +  20002^  —  500002  +  100000  =  0. 

We  find,  on  trial,  that  0,  1,  and  2  give  positive  results ;  3  and  4 
give  negative  results;  5,  and  all  numbers  above  5,  give,  again,  positive 
results.  Hence,  2  and  4  are  the  tenths  of  the  sought  values;  and 
1-2,  and  1'4  are  those  values,  approximatively. 

Now,  if,  in  the  transformed  equation  in  z,  two  values  had  lain  be- 
tween consecutive  numbers,  we  must  have  had  recourse  either  to 
mihute  substitutions,  or  to  the  equation  of  differences.  If,  for  instance, 
the  second  value  had  been  1-26  instead  of  1-4,  the  tenths  would  have 
been  the  same  for  both  values,  and  the  substitution  of  the  natural  num- 
bers in  2  would  have  given  no  change  of  sign. 

To  obviate  a  difficulty  that  might  occur  in  more  than  one  of  the 
transformed  equations,  we  proceed  thus. 

We  take  all  the  functions,  X,  X',  X",  &c.,  and  transform  them  into 
others  in  y,  so  that  the  values  of  y  shall  differ  from  those  of  x,  by  the 
initial  figure,  which,  in  this  case,  is  unity.     We  will  then  have  the 

Y  =  /  +  5y'  +  8/  +  2/  — 5y  +  1, 
Y'  =  by'  +  20y«  +  24/  +  4y  —  5, 
Y"  =  4/-M8/-f24y_10, 
Y'".==- 42/ -252^  +  90, 
Y'^  ^  —  2880^  +  960, 
Y"  =  — ,  a  constant. 
41* 


486  s»urm's  theorem. 

After  having  found  Y,  as  indicated,  we  may  get  its  derivative,  Y', 
and  then  divide  Y  by  Y',  and  proceed  as  we  did  when  getting  the 
functions,  X,  X',  X",  &c.  But,  in  genei-al,  the  better  method  is  to 
transform  the  functions  in  x  into  others  in  y,  so  tliat  the  values  of  y 
shall  be  less  than  those  of  x,  by  the  initial  figure  of  the  sought  value. 


msformed  functions  in  z  are : 

=  z5  ^  50z^  +  800z»  +  2000z^  —  50000z  +  100000, 
'  =  5^"  +  2002^  +  24002^  +  4000z  —  50000, 


The  transformed  functions  in  z  are : 

Z  = 

Z'_.„    . , ,    

Z"  =  \z^  +  ISOz^  +  2400z_  10000, 
Z'"  =  _42z2  — 2520z  +  9000, 
Z"  =  —  2880z  +  9600, 
Z'  =  — ,  a  constant. 

And  we  have  the  following  results  : 


z, 

Z', 

Z", 

Z'" 

Z", 

Z\ 

When  z  =  0 

+ 

— 

— 

+ 

+ 

— ,  three  variations, 

"      z  =  l 

+ 

— 

— 

+ 

+ 

a              u 

"     z  =  2 

+ 

— 

— 

+ 

+ 

cc             a 

"     z  =  S 

— 

— 

— 

+ 

+ 

— ,  two  variations, 

«     z  =  4 

— 

— 

+ 

— 

— 

___          U                 li 

"     z  =  5 

+ 

+ 

+ 

— 

— 

— ,  one  variation. 

We  need  go  no  further  in  the  substitution,  since  5  gives  the  same 
number  of  variations  as  +  oo.  We  see  that  there  is  a  variation  lost 
between  2  and  3,  and  another  between  4  and  5.  Hence,  2  and  4  are 
the  tenths  in  the  required  values,  as  we  before  found.  Now,  if  there 
had  been  two,  or  any  number  of  values  having  the  same  initial  decimal 
figure,  3,  for  example,  there  would  have  been  as  many  variations  lost 
or  gained  as  there  were  values  having  the  same  initial  decimal. 

494.  We  see,  then,  the  two  great  advantages  of  Sturm's  theorem  in 
finding  the  irrational  values  over  the  other  three  processes  described : 
1st.  When  an  equation  comprises  an  even  number  of  values  between 
two  consecutive  numbers,  it  enables  us  to  detect  these  values  without 
minute  substitutions,  or  the  employment  of  the  equation  of  diiferences. 
2d.  When  two  or  more  values  have  the  same  initial  decimal  figure,  it 
enables  us  to  tell  the  exact  number  of  those  values.  If  we  add  to  these 
two  advantages  the  one  first  mentioned,  that  of  detecting  directly  the 


Sturm's  theorem.  487 

number  of  real,  and,  consequently,  the  number  of  imaginary  values,  we 
can  see  how  important  the  theorem  is. 

To  get  the  hundredths,  the  functions  in  z  may  be  transformed  into 
others  in  s,  so  that  the  values  of  s  shall  differ  from  those  of  z  by  the 
initial  decimal  figures ;  in  this  case,  2  and  4.  Then,  again,  transform 
the  functions  in  s  into  others  in  lo,  so  that  the  values  of  w  shall  be  ten 
times  greater  than  those  of  s.  We  will  then  have  a  series  of  functions, 
W,  W,  W",  W",  &c.,  in  which  we  may  substitute  the  natural  num- 
bers, 0,  1,  2,  &c.,  until  we  get  as  many  variations  as  -j-  oo  gives  in  the 
series.  The  least  of  the  consecutive  numbers,  between  which  a  loss  of 
variation  occurs,  will  be  hundredths  in  the  sought  value.  We,  of 
course,  will  have  two  series  of  functions  in  to,  the  one  corresponding  to 
2,  as  the  initial  figure  of  decimals,  and  the  other  to  4. 

495.  Since  there  were  three  real  values  in  the  equation,  x^  —  2x'  — 
2x^  +  4  =  0,  and  we  have  found  but  two  positive  values,  the  other 
must  be  negative.  To  determine  this  negative  value,  let  us  resume 
the  functions, 

X  =  x5  —  2x'  — 2x^  +  4  =  0, 

X'  =  5a;''  —  Qx^  —  4:X, 

X"  =  4x»  +  6x''  — 20, 

X'"  =  —  42a;2  —  168x  +  300, 

X''  =  —  2880x  +  3840, 

X"  :=  — ,  a  constant. 

2,  —  3,  &c.,  in  the  foregoing  series,  we  will 

X",  x^ 

+  — ,  three  variations, 
+  — ,  three  variations, 
+     — ,  four  variations. 

And,  since  —  2  gives  the  same  number  of  variations  as  —  oc,  it  is 
the  superior  negative  limit.  And,  since  a  gain  of  variation  occurs  be- 
tween —  1  and  —  2,  the  entire  part  of  the  negative  value  is  —  1.  In 
this  case,  we  know  that  there  is  but  one  negative  value ;  we  may  then 
proceed  at  once  to  find  the  decimal  part  of  this  value  by  the  usual 
process  for  irrational  values,  without  forming  the  functions,  Y,  Y', 
Y";  &c. 


Substituting  0,  — 

1  _2,_3,& 

have 

X,  X',  X",  X' 

When,     x=       0 

+    0     -     + 

"         x  =  —l 

+   +    —     + 

"         x=-2 

-  +    —     + 

488  Sturm's  theorem. 

.     The  transformed  equation  in  ^  is 

2^'  — 5/ +  8/  — 6^^  +  3y  +  3  =  0, 

and  that  in  z. 

z5_  502^  +  SOO^^'  —  6000/  +  30000z  +  300000  =  0. 

A  change  occurs  between  4  and  5.  Hence,  4  is  the  tenths  of  the 
sought  value,  and  we  have  x  —  — 1-4  for  the  approximate  value. 
Now,  diminish  the  values  in  the  equation  in  z  by  4,  and  the  trans- 
formed equation  in  s  will  be 

8^  —  208"  +  1760s='  — 21040s2  +  130480s  +  21536  =  0, 

and  that  in  w  will  be, 

Mj5— 200wHl76000w;='— 21040000«;2-^1304800000w+2153600000=0. 

A  change  of  sign  occurs  between  the  results  after  the  substitution 
of  —  1  and  —  2.  Hence,  1  is  the  hundredths  of  the  negative  value. 
And  we  have,  for  a  second  approximation,  x  =  —  141. 

Hemarh. 

496.  There  is  one  point  of  considerable  importance  in  the  demon- 
oration,  which  deserves  to  be  attended  to.  It  has  been  shown  that  a 
variation  is  lost  or  gained  between  X  and  X',  every  time  that  X  be- 
comes equal  to  zero,  or  that  x  passes  a  value.  It  might  be  asked,  then, 
Why  not  confine  the  substitutions  to  X  and  X'  ? 

It  is  to  be  observed,  that  a  change  of  variation  only  takes  place  be- 
tween X  and  X'  when  very  minute  substitutions  are  made.  For,  in 
the  demonstration  of  the  fourth  principle,  we  supposed  ii  to  be  indefi- 
nitely small.  Now,  if  we  substitute  a  number,  p,  which  will  make  the 
signs  of  X  and  X'  contrary,  and  again  substitute  another,  jj',  there  being 
two  values  of  X,  and  none  of  X',  between  p  andp',  then  X  and  X'  will 
still  be  affected  with  contrary  signs.  So,  that,  between  p  and  p',  there 
will  be  no  change  of  variation,  though  there  are  two  real  values.  But, 
if  one  value  of  X'  lies  between  p  and  p',  there  will  be  one  variation 
lost,  and  but  one.  For,  whilst  X  changes  its  sign  twice,  X'  will  change 
its  sign  once. 

Take  X  =  ex''  —  5u;  ^-  1 ;  then,  X'  =  12x  —  5. 


stuem's  theorem.  489 

There  is  a  variation  between  X  and  X'  when  x  =  0,  and  this  is 
changed  into  a  permanence  when  x  =  1.  There  are  two  values  of  X 
between  0  and  1,  and  one  of  X'. 

Again,  if  there  were  three,  five,  or  any  odd  number  of  values  of  X 
passed  over,  and  none  of  X',  there  would  be  one  variation  lost  or 
gained,  and  but  one.  But,  if,  at  the  same  time,  an  odd  number  of 
values  of  X'  were  passed  over,  there  would  be  no  change  of  variation. 
It  is  plain,  then,  that  the  loss  and  gain  of  variation  between  X  and  X' 
will  only  correspond  to  the  number  of  real  values,  when  the  substitu- 
tions are  so  minute  that  no  value  of  either  X  or  X'  is  contained  be- 
tween them. 

Scholium. 

497.  Whenever  any  function  is  constantly  positive  for  all  values  of 
X,  we  need  not  form  any  other  functions,  but  only  count  the  number 
of  variations  given  by  +  oo,  and  —  oo,  in  the  constantly  positive  func- 
tion, and  in  the  functions  which  precede  it.  For,  if  we  formed  the 
succeeding  functions,  they  would  give  tlie  same  number  of  variations 
for  -(-  CO  that  they  would  for  —  oo.  Hence,  the  difference  between 
the  number  of  variations  of  these  functions  must  always  be  zero.  To 
show  this,  we  take  for  granted  that  a  function,  which  always  remains 
positive  for  all  values  of  x,  must  be  of  an  even  degree,  since  it  will 
contain  only  imaginary  values.  The  next  succeeding  function  may 
have  its  leading  term  either  positive  or  negative,  but  the  nest  must  be 
negative,  since  any  value  that  reduces  the  function,  consecutive  with  the 
positive  one  to  zero,  must  cause  the  adjacent  functions  to  be  affected 
with  contrary  signs.  All  the  functions  of  an  odd  degree  may  be  either 
positive  or  negative,  but  those  of  an  even  degree  must  be  alternately 
positive  and  negative. 

Let  x^  +  mx^  +  &c.,  be  the  constantly  positive  function.  We  will 
then  have  the  series, 

+  x^  +  mx^  +  &c.  =  X', 


±  n^K^  +  &c. 

=  x^n 

—  px'^  —  &c. 

=  X^+^ 

dr  rx'^  +  &c. 

=  X^, 

+  sx^  +  &c. 

=  x^, 

±tx  -\-  &c. 

=  x^^ 

—  A,  a  constant, 

=  x-n 

490  Sturm's  theorem. 

Now,  suppose  the  ambiguous  sign  to  be  plus  througliout,  we  will 
have  for  -f  oo,  these  results, 

+  H h  +  H J  3  variations, 

and  for  —  oo,  these  results, 

-\ 1 ,  3  variations. 

Next,  suppose  the  ambiguous  sign  to  be  minus  throughout,  then, 
—  00  will  give 

+  -\ H  +  H J  3  variations, 

and  +  oo  will  give 

H 1 ,  3  variations. 

We  will  find  like  results  when  some  of  the  ambiguous  signs  are 
taken  as  positive,  and  the  rest  as  negative. 

GENERAL  EXAMPLES. 

1.  Given,  x^  —  x^  —  7  =  0,  to  find  x. 

X  =  x''  —  x'  —  7  =  0, 
X'  =  Sx''  —  2x, 
X"  =  2x  +  63, 

X'"  =  — ,  a  constant. 

+  oo  gives  +  +  H ,  1  variation, 

—  oo  gives (-  ^ ,  2  variations. 

Hence,  one  real  value,  and  x  =  2-310. 

2.  Find  one  value  of  x  in  the  equation,  x'^  —  2x'  -f  x^  —  5  =  0. 

Ans.  X  =  2-076. 
X  =  x*  —  2x^-j-x^  —  5  =  0, 
.  X'  =  4x^  —  Qx^  +  2x, 
X"  =  x'  —  x  +  20, 
X'"  =  2a;  —  1, 
X'^  =  —  A,  a  constant. 

4-  00  gives  +  +  +  H ,  1  variation, 

—  oo  gives  -\ 1 ,  3  variations. 

Hence,  two  real  values. 


DESCARTES      RULE.  « 

3.  Given,  a;''  —  a;  —  7  =  0,  to  find  x.  Ans.  x  =  2-086. 

X  =  a;='  —  a;  —  7  =  0, 
X'  =  3x^—1 
X"  =  2x  +  21, 
X'"  =  —  A. 

+  00  gives  +  +  -I ,  1  variation, 

—  ex  gives 1 ,  2  variations. 

Hence,  one  real  value. 

4.  Given,  x^  —  x^  +  7  =  0,  to  find  x.  Ans.  a:  =:  — 1-63. 

X  =  a:^  — a;*  +  7  =  0, 
X'  =  3x''  — 2a:, 

X"  =  2a;  —  63, 
X'"  =  —  A. 

+  00  gives  +  4-  H J  1  variation, 

—  oo  gives 1 ,2  variations. 

Hence,  one  real  value. 


DESCARTES'    RULE. 

498.  ^4.?!.  equation  cannot  have  a  cp-eater  number  of  negative  values 
than  there  are  permanences  of  sign  from  +  to  +,  or*  from  —  to  — ; 
nor  can  it  have  a  greater  number  of  positive  values  than  there  are 
variations  of  sign  from  +  to  — ,  or  from  —  to  -\- . 

When  the  adjacent  terms  of  an  equation  are  affected  with  the  same 
sign,  a  permanence  is  said  to  exist  between  them ;  and  when  they  are 
affected  with  contrary  signs,  there  is  a  variation  between  them. 

In  applying  Descartes'  rule  to  an  equation,  every  sign  is  read  twice, 
except  the  first  and  last.  Thus,  in  the  equation,  x*  —  4x^ -f  12a;^  + 
X  —  5  =  0,  there  is  a  variation  between  the  first  and  second  terms,  a 
variation  between  the  second  and  third  terms,  a  permanence  between 
the  third  and  fourth  terms,  and  a  variation  between  the  fourth  and 
fifth  terms.     In  all,  there  are  three  variations  and  one  permanence. 


492  DESCARTES'    RULE. 

When  a  term  is  missing  from  an  equation,  it  must  be  supplied  witli 
tlie  coefficient,  ±  0.  Then,  in  reading  the  signs,  we  must  first  count 
the  variations  and  permanences,  regarding  the  ambiguous  sign  as  posi- 
tive, and  then  count  again,  regarding  the  ambiguous  sign  as  minus. 
Thus,  taking  the  equation,  x^  —  4  =  0,  we  must  write  it  x^  ±0x  — 
4  =  0.  Counting  the  upper  sign,  we  have  a  permanence  between  the 
first  and  second  terms,  and  a  variation  between  the  second  and  third 
terms.  Counting  the  lower  sign,  we  have  a  variation  between  the  first 
and  second  terms,  and  a  permanence  between  the  second  and  third 
terms.  In  whatever  way,  then,  we  read  the  ambiguous  sign,  there  will 
be  one  permanence,  and  one  variation. 

If  there  are  any  number  of  missing  terms,  they  must  be  supplied  in 
like  manner  with  positive  or  negative  zero  coefficients. 

499.  To  demonstrate  the  rule  of  Descartes,  it  is  necessary  to  show 
that  the  multiplication  of  any  equation  by  a  factor,  corresponding  to  a 
negative  value,  will  introduce  into  the  new  equation  at  least  one  more 
permanence  than  existed  in  the  old ;  and  that  the  multiplication  by  a 
factor,  corresponding  to  a  positive  value,  will  introduce  at  least  one 
more  variation. 

1st,  Let  us  take  the  equation,  x"'  —  Px"-'  +  Qx'^-^  +  Rx™""  — 
Sa;""^ .  .  .  .  +  Tx  —  U  ^  0,  and  multiply  by  a  factor,  x  +  a,  corres- 
ponding to  a  negative  value.     The  resulting  equation  will  be 

a;m+l_p|.^n,  +  Qbm->  +  R|x'»-^— S|x»-=' -fTlx"— Ulx  =  0 

-fal   — Pal        +Qa\      +'Ra\  — Sal  +Tal— Ua     ^■' 

Now,  it  is  plain  that,  if  we  suppose  the  coefficients  in  the  upper 
column  are,  throughout,  greater  than  the  corresponding  coefficients  in 
the  lower  column,  there  will  be  the  same  number  of  permanences  in 
the  new  as  in  the  old  equation,  until  we  get  to  the  last  two  coefficients 
(the  coefficients  of  x  and  x°),  we  will  then  have  one  more  permanence 
than  in  the  given  equation.  And,  if  we  suppose  the  coefficients  of  the 
lower  column  to  be  greater  throughout,  there  will  be  a  new  permanence 
between  the  first  and  second  terms,  and  the  same  succession  of  signs  in 
the  remaining  terms.  Moreover,  it  is  evident,  that  if  the  lower  column 
sometimes  prevailed,  and  sometimes  did  not,  there  might  be  more  than 
one  permanence  introduced.  For  instance,  if  a  were  greater  than  P, 
Pa  less  than  Q,  Ra  greater  than  S,  T  greater  than  Sa,  Ta  greater  than 
U,  there  would  be  but  one  variation  in  the  new  equation. 


DESCARTES'    RULE.  493 

It  is  even  possible  to  change  all  the  variations  into  permanences,  by 
multiplication  by  a  factor  corresponding  to  a  negative  value.  As  an 
illustration,  take  the  equation,  x^  —  2a;*  +  \2x*  —  8x^  +  36x-  —  x  •\- 
15  =  0,  and  multiply  by  cc  +  4.     The  new  equation  will  be, 

x-'  —  2.\x^  +  11\x^—   Slx^ +  36|a;3—     1  j  x^  +  15  Ix  =    0 
+  41     —si       +  48  I      —  32  I      +  144  I     —   4  |    +  60' 

or,         x'  +  2a;«  +  ^x'  +  40^"  +  4x='  +  143x'=  +  llx  +  60  =  0. 

By  this  example  we  see  that  an  equation,  containing  only  variations, 
is  changed  into  another  containing  only  permanences,  by  multiplying 
the  former  by  a  factor  corresponding  to  a  negative  value.  Seven  per- 
manences have  been  introduced  where  none  existed  before.  And,  by 
recurrence  to  equation  (M),  we  see  that  it  is  impossible  to  read  the 
signs  in  any  order,  without  having  one  more  permanence  than  in  the 
given  equation.  Now,  the  given  equation  may  have  been  one  of  the 
first  degree,  m  being  equal  to  one,  and  P,  Q,  K,  S,  and  T  being  equal 
to  zero.  Then,  if  the  sign  of  U  were  positive,  there  would  be  one 
negative  value,  and  conversely.  The  new  equation  (after  multiplica- 
tion by  x  +  ci)  would  be  of  the  second  degree,  and  would  contain,  at 
least,  one  more  permanence  than  the  old.  And,  by  multiplying  this 
new  equation  by  another  factor  corresponding  to  a  negative  value,  we 
would  introduce,  at  least,  one  more  permanence.  And  so,  by  contiuu 
ing  the  process,  it  could  be  shown  that,  whatever  might  be  the  degree 
of  the  equation,  the  number  of  permanences  must  always  be  equal  to, 
or  exceed  the  number  of  negative  values. 

500.  2d.  By  a  similar  course  of  reasoning,  we  could  show  that  the 
the  multiplication  by  a  factor  corresponding  to  a  positive  value  would 
introduce,  at  least,  one  variation  ;  or,  in  other  words,  that  the  number 
of  positive  values  can  never  exceed  the  number  of  variations. 

Take,  as  an  illustration,  the  equation, 

a;6  _  4x5  +  i2x'  +  8.x3  +  SO.r^  —  x  -f  15  =  0, 
and,  multiply  it  by  the  factor,  x  —  2,  the  resulting  equation  will  be, 

a;'  —  6x«  -f  20x5  —  lOx''  +  14x'  —  61x2  -f  17x  —  30  =  0, 
and  has  gained  three  variations. 

501.  It  is  plain  that  the  preceding  reasoning  has  been  on  the  sup- 
position that  a  was  a  real  value,  otherwise  we  could  not,  in  equation 

42 


494  DESCARTES'    RULE. 

(M),  have  instituted  any  comparisons  betweeij  a  and  —  P,  —  Pa,  and 
4-  Q,  +  R«>  and  —  S,  &c.  In  case,  then,  that  there  are  imaginary 
values  in  an  equation,  the  rule  of  Descartes  only  points  out  limits,  be- 
yond which  the  positive  and  negative  values  cannot  go.  But,  when 
the  equation  contains  only  real  values,  the  number  of  positive  values 
will  be  exactly  equal  to  the  number  of  variations,  and  the  number  of 
negative  values  exactly  equal  to  the  number  of  permanences.  To  show 
this,  let  m  =  degree  of  the  equation,  ?i  =  number  of  real  negative 
values,  p  =  number  of  real  positive  values,  h  =  number  of  variations, 
b'  =  number  of  permanences.  A  slight  inspection  will  show  that,  in 
case  of  real  values,  b  +  b'  =  7n ;  and  we  know  that  n  +  p  =  m. 
Hence,  b  -\-  b'  =  n  -\-  p.  But,  since  n  cannot  exceed  b',  and  p  cannot 
exceed  b,  we  must  have  n  =  b'.  For,  if  ra  <^  6',  then  necessarily 
p^  b,  which  cannot  be.  In  like  manner,  we  could  show  that  we 
must  havej9  =  5. 

Corollari/. 

502.  In  case  of  there  being  one  or  more  missing  terms  in  an  equa- 
tion, the  rule  of  Descartes  will  enable  us  to  detect  imaginary  values. 
We  have  only  to  supply  the  missing  terms  with  plus  or  minus  zero 
coefficients,  count  the  variations  and  permanences  when  the  upper  sign 
is  taken,  and  then  again,  when  the  lower  sign  is  taken.  If  there  be 
any  discrepancy  in  the  results,  there  will  be  imaginary  values.  Thus, 
take  cc^  +  4  =  0 ;  supplying  the  missing  term,  we  have  x^  ±  Ox  -^  4: 
=  0.  The  upper  sign  taken  in  connection  with  the  other  two,  gives 
two  permanences,  whilst  the  lower  gives  two  variations.  These  dis- 
crepant results  indicate  imaginary  values. 

Take  x*-\-Sx^^4:  =  0,  then,  x*  db  Ox'  +  Sx^  ±  Ox  —  4  =  0. 

In  one  case,  we  have  one  variation  and  three  permanences ;  in  the 
other,  three  variations  and  one  permanence.  The  difference  in  the  two 
readings  again  indicates  imaginary  values. 


GENERAL   EXAMPLES. 

1.  What  are  the  values  in  the  equation,  x^  —  2x^  —  7x  -|-  1  =  0  ? 

2.  What  are  the  values  in  the  equation,  x"  —  7x  -f  1  =  0  ? 

3.  What  are  the  values  in  the  equation,  x''  —  6x*  -f  7x^  —  Sx*  -|- 
9x«4-  10x2  —  11x4-  12  =  0? 


ELIMINATION    BETWEEN    TWO    EQUATIONS.  495 

4.  Wliat  are  the  values  in  the  equation^  x"  —  1  =  0? 

5.  What  are  the  values  in  the  equation,  x^  —  1  =  0? 

6.  "What  are  the  values  in  the  equation,  x^  +  1  =  0  ? 

7.  What  are  the  values  in  the  equation,  x^  +  4a;^  +  6x^  +  4a;  + 
1  =  0? 


ELIMINATION  BETWEEN  TWO  EQUATIONS 
OF  ANY  DEGREE. 

503.  When  one  quantity  depends  upon  another  for  its  value,  it  is 
said  to  be  a  function  of  the  quantity  upon  which  it  depends.  Thus,  in 
the  equation,  y  =  'Zx  —  4,  ^  is  a  function  of  x,  because  every  change 
in  the  value  of  x  will  produce  a  corresponding  change  in  the  value 
ofy. 

The  mathematical  symbol  to  designate  a  function  may  be  F,  or/, 
or  the  Greek  letter,  $.  Thus,  to  indicate  that  i/  is  a  function  of  x,  we 
may  employ  the  notation,  y  =.  F(:c),  or  y  =/(x'),  ox  y  ^  4'(a:)-  The 
second  member  may  contain  constants,  as  well  as  the  variable,  x.  Thus, 
3/  is  a  function  of  x  in  the  foregoing  equation,  y  =  2.x  —  4,  the  con- 
stants being  2  and  —  4. 

504.  The  most  general  form  of  an  equation  of  the  m*"  degree  be- 
tween two  variables,  x  and  y,  is, 

cc"  -f  Bx"-'  +  Cx---^  +  Dx"-^  +  Ex"-^ +  U  =  0 ; 

in  which  B,  C,  D,  &c.,  are  functions  of  ?/. 

B  is  supposed  to  be  of  the  first  degree  in  y,  and  of  the  form, 
a  -\-hy. 

C  is  of  the  second  degree  in  y,  and  of  the  form,  c  -\-  cly  -\-  ey^. 

D  is  of  the  third  degree  in  y,  and  of  the  form,  f -\-  gy  ■\-  hy^  +  1y^- 

E  is  of  the  fourth  degree  in  y,  &c.  &c. 

U  is  of  the  m}^  degree  in  y,  and  of  the  form,  u  -f  my  + +!/°7 

and  it  does  not  contain  x. 

505.  The  equation  is  said  to  be  complete  when  x  enters  into  all  the 
terms  but  the  last,  y  into  all  the  terms  but  the  first,  and  when,  also, 
the  sum  of  the  exponents  of  x  and  y  in  each  term  is  equal  to  m. 


496  ELIMINATION    BETWEEN    TWO    EQUATIONS. 

506.  Eliminatioa  between  equations  of  a  degree  higher  than  the 
first  is  usually  effected  by  means  of  the  greatest  common  divisor.  This 
method  of  elimination  has  already  been  explained  (Art.  214),  but  we 
propose  to  demonstrate  the  process  more  rigorously,  in  two  ways, 

1.  Let  A  =  0,  and  B  =  0,  be  the  proposed  equations  containing 
both  X  and  y. 

Now,  if  we  knew  beforehand  a  value  m  of  y,  that  was  common  to 
the  two  equations,  A  =  0,  and  B  =  0,  and  substituted  this  value  in 
them,  the  new  equations.  A'  =  0,  and  B'  =  0,  would  contain  only  x 
and  constants.  Now,  if  n  be  a  value  of  a;,  in  the  equation.  A'  =  0, 
its  first  member  must  be  divisible  by  x  —  n,  and  it  is  plain  that  the 
given  equations  would  not  be  simultaneous  unless  n  would  also  satisfy 
the  equation,  B'  =  0  (Art.  207).  Hence,  x  —  n  must  also  be  a  divi- 
sor of  the  equation,  B'  =  0.  We  see,  then,  that  the  hypothesis  of  a 
common  value  in  y  results  in  the  condition  of  a  common  divisor  in  x. 
Conversely,  if  we  can  force  the  given  equations  to  have  a  common 
divisor  in  x,  they  must  have  a  common  value  in  y.  It  is  upon  this 
principle  that  we  seek  for  a  common  divisor  between  the  first  members 
of  the  equations,  A  =  0,  and  B  =  0,  and  continue  the  process  until 
we  get  a  remainder  freed  from  x.  It  is  plain  that,  if  we  place  this 
remainder  equal  to  zero,  and  substitute  the  value  of  y,  found  from  it  in 
the  last  divisor,  it  will  be  an  exact  divisor,  and  will  be  the  one  sought. 

From  the  foregoing  reasoning  it  is  evident,  that  if  it  be  absurd  to 
place  the  remainder,  freed  from  cr,  eqvxal  to  zero,  the  given  equations 
are  not  simultaneous. 

507.  2d.  Let  the  successive  quotients,  in  the  process  of  dividing  A 
by  B,  B  by  the  remainder,  &c.,  be  designated  by  Q,  Q',  &c.,  and  the 
successive  remainders  by  R,  R',  &c.     Then  we  will  have. 

A  =  BQ  4-  R,  (M) 

B  =  RQ'  +  R',  (N) 

•       R  =  R'Q"  +  R",  (0) 

&c.         &c. 

Now,  since,  by  hypothesis,  A  =  0,  and  B  =  0,  equation  (M)  will 
give  R  =  0.  And,  since  B  =  0,  and  R  =  0,  equation  (N)  will  give 
R'  :=  0.  From  this  we  see  that  we  have  a  right  to  equate,  with  zero, 
that  remainder  which  is  freed  from  cc,  and  contains  only  y. 

508.  The  above  series  of  equations  show,  moreover,  that  if  the  re- 
mainder, which  is  freed  from  a-,  and  the  preceding  divisor  be  placed 


OF    ANY    DEGREE.  497 

equal  to  zero,  the  system  of  values  so  found  will  satisfy  tlie  given  equations. 
For,  if  E."  be  that  remainder,  and  R'  the  preceding  divisor,  when  E,", 
and  R'  are  equated  with  zero,  equation  (0)  shows  that  R  also  =  0. 
And  R  and  R'  being  equal  to  zero,  from  (N)  we  get,  B  =  0.  And, 
since  B  =  0,  and  R  =  0,  equation  (M)  shows  that  we  will  also  have 

A=:0. 

Hence,  the  values  of  ?/,  found  by  placing  the  last  remainder  equal  to 
zero,  may  be  substituted  in  the  preceding  divisor,  in  order  to  deduce 
the  corresponding  values  of  x.  The  importance  of  this  remark  consists 
in  the  fact,  that  the  preceding  divisor  is  of  a  lower  degree  than  either 
of  the  original  equations,  and,  therefore,  more  readily  solved. 

509.  The  reasoning  in  Art.  507  proceeds  upon  the  supposition  that 
the  successive  quotients,  Q,  Q',  Q",  are  all  finite,  and  then,  of  course, 
the  successive  products,  BQ,  RQ',  R'Q",  &c.,  will  all  be  zero,  when 
B  =  0,  R  =  0,  R'  =  0,  &c.  But,  if  any  of  these  quotients  be  frac- 
tional in  form,  it  may  happen  that  the  value  of  y,  found  from  placing 
the  last  remainder  equal  to  zero,  will  reduce  the  denominator  of  the 
preceding  divisor  to  zero  also.  In  that  case,  we  would  have  the 
product  of  zero  by  infinity,  which  is  indeterminate.*     Suppose,  for 

4 
example,  R  =  y  —  1,  Q  =  —^ .      Then,  A  =  BQ  -f  R  becomes 

A  =  ^(-A—\  +  y  —  1 :  or,  when  R  =  0,  A  =  B^  =  B  oc ;    or 

V  — y  0 

(since  A  and  B  are  zero),  0  ^  0  oo,  which  may,  or  may  not  be,  a  true 
equation. 

510.  To  avoid  the  fractional  form  of  quotient,  it  may  sometimes  be 
necessary  to  multiply  the  dividend  by  some  function  of  y,  though  this 
multiplication  may  possibly  introduce  some  foreign  values  into  the 
equation,  as  will  be  shown  more  fully  hereafter. 

511.  We  will  now  illustrate  the  foregoing  principles  by  a  few 
examples. 

*  Let  0  00  :iz  A,  dividing  both  members  by  0,  we  get  oo  ==  — -  =:  oo,  a  true 

equation.  And  this  will  evidently  be  true  when  A  =  1,  5,  10,  20,  or  anything 
whatever. 

42*  2a 


49S  ELIMINATION    BETWEEN    TWO    EQUATIONS 

Let  a;2  +  J/''  —  8  =  0,  and  2a;  —  3y  +  2  =  0. 

4 

4x2+4/— 32  \2x  —  ^y  +  2 
4ic^+6.ry+4.x     2a:  +  (By  —  2)  =  Quotient. 
1st  Kemainder  =  2(3?/— 2)a;  +  4/— 32 

2(3y— 2)a:— 9/+12^— 4 
2d  Remainder  =  13^2—12^—28=0. 

From  which  we  get  y  =  2,  and  y"  =  —  i| ;  and  these,  when  sub- 
stituted in  either  of  the  given  equations,  give  x'  =  2,  and  x"  =  — 1|, 

Let  x^  —  /  —  7  =  0,  and  a;  —  y  —  1  =  =  0.  Combining,  we 
will  get  a;^  +  (^  +  \)x  +  (/  +  y)  +  (^  +  1)  for  a  quotient,  and 
/+  t/  —  2  for  a  remainder. 

The  equation,  formed  by  placing  the  remainder,  freed  from  x,  equal 
to  zero,  is  called  i\i.e  final  equation.  In  this  example,  the  final  equa- 
tion, y^  +  y  —  2  =  0,  gives  %f  =  1,  y"  =  —  2.  And  these  values 
for  y,  when  substituted,  give  x'  —  2,  and  x"  =  —  1. 

Let  x"  —  Syx^  +  3/a;  —  5a;*  -f  lO^a;  +  6a;  —  /  —  5y^  —  Qy  =  0, 
and  a;'  —  6yx^  +  Sy^x  —  x  —  4/  -|-  y  =  0.  Then  the  first  quotient 
is  -|-  1,  and  first  remainder   (2y  —  5)x^  —  5y^x  -\-7x  +  IQyx  +  By^ 

—  5/  —  7y.  Preparing  the  last  divisor  for  division  by  multiplying  by 
(2y  —  5)^,  we  have 

x^—Syx^-^SiJ'x—x—iy^+T/ 

(2^-5)" 

(2y-5)^x'-20y^x^+32y''x+im>/^x+lOy^a^-12byx'^-imy^x+20yx-25x-16y^+80y*-96y^-20y<^+25]/, 

and  this,  when  divided  by  (2y  —  5)x^  —  by^x  +  7x  +  lOyx  +  3/ 

—  5/,  gives  as  a  quotient,  (2y  —  5)x  —  5/  -f  15y  —  7,  and  as  a  re- 
mainder y*x  —  10/a;  +  B5y^x  —  bOyx  +  24a;  —  y^  +  lOy*  —  35^'  + 
50/  —  24y.  And,  by  factoring  this  remainder,  we  get  (/  —  10/  -f- 
35/  —  bOy  +  24)x  —  3/(/  —  10/  +  35/  —  50y  +  24) ;  or,  (a;  —y) 
(/  —  lOy^  +  35^2  —  50?/  +  24).  Rejecting  the  factor,  x  —  y,  as 
leading  to  arbitrary  values,  we  have  the  final  equation,  y*  —  lOy^  -\- 
35/  —  50y  +  24  =  0.  This  equation,  when  solved  by  the  process 
of  divisors,  gives  y'  =  1,  /  =  2,  /"  =  3,  and  y'"^  =  4.  And  these, 
when  substituted,  give  x'  =  3,  x"  =  5,  x'"  =  5,  a;'^  =  7. 

512.  Two  things  are  suggested  by  this  example.  1st.  May  not  the 
multiplication  by  the  factor,  (2y  —  by,  involving  One  of  the  unknown 
quantities,  have  introduced  foreign  values ;  that  is,  values  which  did 
not  enter  the  given  equations  ?     2d.  x  —  y  being  a  common  factor  to 


OP    ANY    DEGREE.  499 

the  remainder,  is  found  to  be  also  common  to  both  the  given  equations. 
How  are  such  factors  to  be  treated  ?  We  will  examine  these  subjects 
separately. 

513.  1st.  In  regard  to  foreign  values,  we  have  this  simple  test. 
Place  the  multiplier  equal  to  zero,  find  the  value  for  y,  and,  from  either 
of  the  given  equations,  the  corresponding  values  of  .r ;  if  these  values 
be  the  same  as  some  of  those  found  from  the  final  equation  for  y,  with 
the  corresponding  values  of  x,  then  the  system  of  common  values  in 
both  X  and  y  must  be  rejected.  In  the  example,  placing  {2.y  —  5)* 
^  0,  we  get  i/  =  I,  a  value  difi'eient  from  those  before  found.  Hence, 
the  multiplier  has  not  introduced  a  foreign  value.  But,  if  the  multi- 
plier, placed  equal  to  zero,  had  given  us,  for  example,  y  =  \,  with  the 
corresponding  x  =  3,  then  this  system  of  values  must  be  rejected. 

514.  How  are  factors  to  be  treated  which  are  common  to  both  of  the 
first  members  of  the  given  equations  ?  There  may  be  three  cases,  but 
all  lead  to  arbitrary  values.  1st.  The  common  factor  may  be  a  func- 
tion of  X  only,  fix).  2d.  It  may  be  a  function  of  y  only,  f{y).  3d. 
It  may  be  a  function  of  both  x  and  y,  f(x,  y). 

When  the  common  factor  is  /(x)  only,  x  will  have  determinate 
values,  and  y  indeterminate.  For,  by  placing  f(x)  =  0,  we  will  get 
true  values  for  x ;  but,  when  f(x)  =  0,  both  equations  will  be  satisfied, 
whatever  values  y  may  have.  Take  the  equations,  (a  -\-  hx')  (x?y  + 
2fx  —  ay)  =  0,  and  (a  -f  hx)  (Axhf  —  2yx  +  my"-)  =  0.      Placing 

a  +  Ja;  ^  0,  we  get  x  = ,  and  both  equations  will  be  satisfied 

when  a  +  hx  =  0,  whatever  may  be  the  values  of  y.  Hence, /(a;)=0 
gives  x  determinate,  and  y  indeterminate.  In  like  manner,  a  common 
factor,  f(y),  would  give  y  determinate,  and  x  indeterminate.  In  the 
the  third  case,  the  common  factor,  /(x,  y)  =  0,  will  satisfy  both  equa- 
tions ;  but,  since  we  have  a  single  equation,  f(x,  y)  =  0,  containing 
two  unknown  quantities,  the  values  of  both  x  and  y  must  be  indeter- 
minate. Thus,  take  the  equations,  (2x  —  4ty)  (a?  -}-  ly  —  5)  =  0, 
and  (2x  —  4y)  (xy  —  1x^  +  bx^  —  7/)  =  0.  It  is  plain  that  both 
equations  will  be  satisfied  when  2x  —  4y  =  0 ;  but  the  equation, 
2x  —  4y  =  0,  will  give  indeterminate  values  for  both  x  and  y. 

Hence,  we  conclude  that,  when  the  given  equations  contain  a  com- 
mon factor,  it  must  be  divided  out.  For,  a  common  factor,  /(x),  would 
give  determinate  values  for  x,  but  indeterminate  for  y;  a  common 
factor,  f(y),  would  give  determinate  values  for  y,  and  indeterminate 


500  ELIMINATION    BETWEEN    TWO    EQUATIONS 

for  x;  and  a  common  factor, /(a;,  ^),  would  give  x  and  y,  both  inde- 
terminate. 

515.  Take  the  equations,  (x  —  1)  (x^  —  2xy  -fa;  —  2)  =  0,  and 
(x^  —  XT/  —  2)  (x  —  1)  =  0.  Suppressing  x  —  1,  we  have  x^  —  2xy 
+  X  —  2  =  0,  and  x^  —  xi/  —  2  =  0,  from  which  we  get  y  =  1,  and 
x  =  2. 

Take  the  equations,  (2y  —  6)  (xy  +  5a-?/  —  y  -f  1)  =  0,  and 
(2i/  —  6)  (xy  -f-  4y  —  4)  =  0.  Suppressing  2y  —  6,  and  combining 
the  resulting  equations,  we  get  IQy^  —  53j/  -f  37  =  0  for  the  final 
equation.     From  which,  ?/'  =  1,  y  =  ^l,  x'  =  0,  x"  =  —  ||. 

Take  the  equations,  (2x  —  7y)  (xi/  —  y  -f  5a;^  —  5a;)  =  0,  and 
(2a;  —  7y)  (xi/  -f-  7a;  —  7  — y)  =  0.  Removing  the  common  foctor, 
we  get,  after  combination,  5a;^  — 12  -f  7  =  0  for  the  final  equation. 
From  which,  x'  =  |,  x"  =  1,  and,  by  substitution,   y'  =  —  7,  and 

516.  When  the  first  members  of  the  given  equations  can  be  resolved 
into  factors  of  the  first  degree,  or  of  a  low  degree,  the  elimination  will 
be  greatly  facilitated. 

Take  the  equations,  (x  —  1)  (?/x  —  3)  (a;^  —  2xy')  —  0,  and  (^x  — 
2a;)  (x^  —  2^)  =  0. 

These  equations  can,  obviously,  be  satisfied  when 

X —    1=01         when  a;  —    1  =  01         when  ya; —    3  =  01 
and  yx  — 2a;  =  01^    ''and     x^— 2y  =  o|*^   -^  and     i/x  —  2x  =  o\^   '' 

when  7/x —  3=0  I        when  x^ — 2xy=0  I        when  x^ — 2xy=0  j 
and     x'—2y=0\^Knd     x^—  2y=0 1^    \nd     i/x—  2x=o\^   ^' 

From  (A)  we  get  the  system  of  values,  ic  =  l,  and  y  =  2. 


"  X  =  1,  and  y  =  i. 

"  X  =  ^,  and  y  =  2. 

"  a;  =  VF,  and  x  =:  ^=z. 

•    ^6 
"  X  =  1,  and  y  =  g. 

«  x'  =  0,  x"  =  4,  and  /  = 


From  (B) 
From  (C) 
From  (D) 

From  (E) 
From  (F) 
y"  =  2. 

An  artifice  will  sometimes  enable  us  to  decompose  the  first  members 
of  the  given  equations  into  their  respective  factors. 


OP    ANY    DEGREE.  501 

Take  the  equation,  cc^  —  lyx,  +  6x  4-  2/*  —  %  +  5  ^  0,  (A),  and 
X*  +  2^a;  +  /  +  6a;  +  6y  +  5  =  0,  (B).  From  (A),  we  get  x^  —  lyx 
+  /  +  6x  —  6j^  +  5  =  0,  or  (x  —  ^)^  +  6(x  —  ^)  +  9  —  4  =  0 ; 
or  (since  tte  first  three  terms  constitute  a  perfect  square),  (x — y+3y 
_4  =  (x— y  +  3/  — (2y  =  (a;  — y  +  5)(x— y  +  l)(Art.  50). 
From  (B),  we  get  (x  +  y^  +  6(x  +  ^)  +  9  —  4  =  0,  or  (x  +  y  +  3)* 
— 4=(x+y  +  3)2  — (2)^  =  (x+3/  +  5)  (x+y  +  1)  (Art.  50).  Hence, 
we  have  (x  —  y  +  5)  (x  —  y  +  1)  =  0,  (A'),  and  (x  +  y  +  5) 
(a;  +  y  +  1)  =  0,  (B').    And  (A')  and  (B')  will  evidently  be  satisfied, 

when  X — ^y-}-5  =  0|  when  x — y  +  5  =  0|         when  x — y-\-\=.^\ 

and     x+y  +  5=0l'^    -^  and     x+y  +  l=Or^  and     x^y^b=^^' 
and  when  x  —  y  +  1  =  0 
and  X  +  y  +  1  =  0 


(Q). 


From  (M)  we  get  the  system  of  values,  x  =  —  5,  and  y  =  0. 
From  (N)       "  "  «  x  =  —  3,  and  3/  =  +  2. 

From  (P)       "  «  «  X  =  —  3,  and  y  =  —  2. 

From  (Q)       "  "  «  x  =  —  1,  and  y  =  0. 

Take  the  equations,  x*  —  3yx  —  3x  +  2/  +  7y  —  4  =  0,  (A), 
and  x^  —  4x  +  x^^  —  4y  =  0,  (B). 

From  (A)  we  get  (by  making  2/  =l  -| ^,  and  adding  and  sub- 
tracting I), 

(-l)"-^(-D+l-(l-l+?)=o 

or,  (-^  - 1  -  I)  -  (f  -  |-)'-:(^-y-4)  (a;-2y+l)  (Art.  50). 

From  (B),  we  get  x(x  —  4)  +  y  (x  —  4)  =  (x  +  y)  (a;  —  4). 
Hence,  we  have  (x  —  y  —  4)  (x  —  2y  +  1)  ^  0. 
and  (x  +  2/)  (a;  —  4)  =  0. 

From  which  we  get  the  equations, 
a;_y_4=0|  a;— y  — 4  =  0  1        cc  — 2y+l=0 


and         x+y  =  Ol^^^'  x  — 4  =  0r^'  x  +  2/  =  0  F^' 

x-2y  +  l  =  0| 

X  — 4  =  0P^ 


502 


ELIMINATION    BETWEEN    TWO    EQUATIONS 


From  (R),  we  get  the  system  of  values,  a;  =  2,  and  y  =  —  2. 
From  (S),       "  "  "  x=i  4,  and  i/  =  0. 

From  (T),       "  "  «  x  =  —  {,  and  ?/  =  +  \. 


From  (U), 


=z  4,  and  y  =  ^. 


Take  the  equations,  x^  +  y^  —  2yx  —  4^  +  4a3  =  0,  (A), 
and  a-y  —  y^  —  6a;y  +  5  +  4y  =  0,  (B). 

From  (A),  we  get  (x  — yf  +  4  (x  — y)  =  0, 
or,  {x—y)(x  —  y+^)  =  (),  (A'). 

And  from  (B),  we  get  (by  adding  and  subtracting  4), 
xY  —  ^xy  +  9  +4y— /  — 4  =  0, 
or,  (xy-%r-(^y-2J  =  ^,  or,  {xy^y-h^  {xy-y-V,=.^,  (B'), 

(A')  and  (B')  give  the  system  of  equations. 


X  —  y  =  0 

and  ccy  +  y  —  5  =  0 

X  —  y  +  4  =  0" 

^y — y  —  1  =  0 


m^ 


x—y  =  Q 
^y—y—l=0 


(H), 


x  —  2/4-4  =  01 
xy-{-y  —  b=(i\ 


(I), 


(K). 


From  (G),  we  get  the  values. 


..^_,  +  41,.^^=:zW2i, 


y' 


—  l  +  v/21 


1  — ^/21 


From  (H),  we  get  the  values, 

— -- 2    J' 
l+v/5 


x/5 


2/"  = 


2    _^ 
1  —  ^/5 


From  (K),  we  get  the  values. 


5+^/29     ,,_  — 5  — -v/29 

^  -        2^3; 

,    3  —  ^29 

y  =+ r, — . 


2     _' 
3  +^/29 


OF    ANY    DEGREE.  503 

517.  One  of  the  equations  only  may  be  capable  of  decomposition 
into  factors. 

The  equations  may  be  of  the  form,  A  =  0,  and  BD  =  0.  We  will 
then  have  two  systems  of  values  j  one  resulting  from  A  =  0  and 
B  =  0,  the  other  from  A  =  0  and  D  =  0. 

Take  the  equations, 

xy  —  Ixy  —  20x  =  0,  (A),  and  {xy  —  5a;  +  3)  (x  —  2)  =  0,  (BD). 
We  get  the  system  of  equations, 

xY  —  Ixy  —  20x  =  0,  (A),  and  cry  —  5x  +  3  =  0,  (B). 
From  which,  x'  =  3,  and  x"  =  |,  y  =  4,  and  y"  :=  —  |. 
(A)  and  (D)  give  xhf  —  Ixy  —  20x  =  0,  and  re  —  2  =  0. 

, . ,        ,       ^    ,     7-^^/209    ,,    i  —  ^lm 

From  which  we  get  x  =z  1,  y  = ,  y  = -r . 

518.  If  any  of  the  successive  remainders  be  capable  of  decompo- 
sition into  factors,  which  are  functions  of  x  or  y,  these  factors  may  be 
placed,  separately,  equal  to  zero,  and  the  deduced  values  of  x  or  y  sub- 
stituted in  the  preceding  divisor. 

For,  let  B,  be  one  of  the  dividends,  R'  the  divisor,  Q  the  quotient, 
and/(x)  X  f{y)  the  remainder. 

Then,  |,  =  Q  +  -^/^l  or  R  =  R'Q  +  f{x)  x  /{y). 

And  this  equation  will  be  satisfied  when  R'  =:  0  and  /(x)  =  0,  or 
when  R'  =  0  &nd/(y)  =  0. 

Take  the  equations,  yx'^  +y^x^  —  x^ —  y^x -^  yx -{- ]/  — 1=0,  (R). 

and  x^  —  y  +  1=0,  (R'). 

Dividing  R  by  R',  we  will  geflfe  remainder,  (x^  +  1)  (y^  —  1) 
Placing  a;*  -f-  1  =  0,  we  get  x  =  ±  -y —  1,  and  this,  substituted  in 
(R),  gives  2/  =  0.     Placing  y^  —  1  =  0,  we  get  y  =  =t  1,  and  these 
values,  when  substituted,  give  x'  =  0,  and  x"  =  db  y/ —  2. 
Hence,  we  have  the  system  of  values, 

a/=+  v/— 1|  x"=—^—  1 1  x'"  =z  0  I  a;'^  =  y/—2\  x^  z=  — ^A^ 
2/  =  0|  y"  =  o\y"'  =  l\      y-  =  — 1|         r  =  —  1 

All  of  which  will  satisfy  the  given  equations. 


504  ELIMINATION    BETWEEN    TWO    EQUATIONS 

519.  When  it  is  necessary  to  multiply  (A)  by  either  a  function  of 
*  or  2/  to  make  it  divisible  by  (B),  we  can  tell  whether  the  multiplier 
has  introduced  foreign  values,  by  combining  it,  placed  equal  to  zero, 
with  B  =  0.  If  any  of  the  values  thus  found  are  the  same  as  those 
resulting  from  the  combination  of  the  given  equations,  we  must  reject 
these  common  values  from  the  solutions  of  the  given  equations. 

For,  Jet  A  =  0,  and  B  =  0 ;  and  suppose  that  (A)  will  not  be  divi- 
sible by  B  until  it  has  been  multiplied  by /(a;).  Then  we  will  have 
A(/x)  =  0,  and  B  ^  0,  which  can  be  satisfied  when  A=0,  and  B  =0; 
or,  when  f(x)  =  0  and  B  =  0. 

Now,  it  is  plain  that,  if  the  combination  of  the  new  equation, 
A.f(w)  =  0,  with  B  =  0,  gives,  among  its  system  of  values,  the  same 
as  given  by /(a;)  =  0,  and  B  =  0,  we  must  reject  the  common  values 
as  having  been  introduced  by  the  multiplication  of /(x). 

Take  the  equations,  x^y  —  2xy  -{-  x^  =  0,  (A),  and  x^y  —  2x^  -\- 
x  =  0,  (B). 

Multiplying  (A)  by  x^,  to  prepare  for  division,  we  get  for  the  final 
equation,  a;^(x'  ^•  2x^  —  5x  +  2)  :^  0.     From  which  we  get,  x  =  0, 

,      ^      „      — 3+VT7     ,„      —3—^17        „,  -. 

X  =  1,  X   = ,  X    = ^ .       The   corresponding 

1  .  /       0      V        6v/T7-26      .„       _(26  +  6^T7) 

values  of  y  are,  v  =  ti,  v   =  — =»  ¥    =  — ^^ =-^ 

"'  ^        38  —  10^17  38  +  10^17 

(M). 

Now,  placing  /(x)  =  0,  and  B  =  0  ;  or,  x"  =  0,  and  cc'  —  2x^  + 
x  =  0,  we  get  x=zO,  and  y  =  § ;  and,  since  these  values  are  the  same 
as  the  first  of  those,  marked  (M),  we  must  reject  them  from  (M),  and 
leave  but  three  values  for  x,  and  three  for  y. 

520.  If  A  be  exactly  divisible  by  B,  the  values  of  x  and  y  will  be 
indeterminate. 

For,  then,  ^  =Q,;  and,  since  A  =  0,  and  B  =  0,  we  will  have 
B 

I]  =  Q,  or  0  =  OQ  .  (P).     Now,  it  k  plain  that  Q  may  be  /(cc),  or 

f{y),oxf(x,ij).     But,  equation  (P)  will  be  satisfied,  whatever  may 

be  the  form  of  Q,  and  whatever  may  be  the  values  of  x  or  y. 

Take  the  equations, 

a;«  _  ^x^y  _  2^''  4-  Zy^x"  +  6xV  —  2/'x  —  Qy^x  +  2f  =  0, 
and  x'  —  2xy  +  y'  =  0. 

We  get  as  a  quotient,  x^  —  xy  —  2x  +  2y,  and  a  remainder  zero, 
and  any  value  whatever  of  x,  with  the  corresponding  or  deduced  value 
of  y,  will  satisfy  both  equations. 


OF    ANY    DEGREE.  505 


Remark. 


The  case  exhibited  in  520,  differs  only  from  that  in  514  in  this 
respect,  there  is  a  common  factor  to  the  two  equations  in  both  instances, 
but  520  does  not  manifest  that  factor. 

The  indeterminate  nature  of  the  given  equations,  when  the  final 
equation  in  y  is  zero  in  both  members,  may  also  be  shown  by  retaining 
the  trace  of  y.  For,  when  we  place  the  zero  remainder  equal  to  zero, 
we  have  Oy  =  0,  or  2/  ^  %■ 

521.  When  the  final  equation  reduces  to  a  constant,  the  value  of  y 
will  be  infinite,  and  the  given  equations  will  be  contradictory. 

For,  then  we  will  have  0^/  =  A,  constant ;  or  ^  =  00,  the  symbol  of 
absurdity. 

Hence,  the  combination  of  the  given  equations  has  led  to  an  absurd- 
ity, and,  therefore,  these  equations  must  be  contradictory. 

Take  the  equations,  y^  —  0?  —  Sx*  +  9  —  Zx  =  0, 

and  y  —  x  —  1=0. 

Combining,  we  will  have,  for  the  final  equation,  10  =  0;  or,  retaining 
the  trace  of  y,  10  -f  Oy  =  0,  or  y  =  —  'gO  =  00, 
The  given  equations  are  not  simultaneous. 

GENERAL  EXAMPLES. 

|/_2xy  — 4x+  a;2  =  0, 
I  y +  a;  — 4  =  0. 

Am.  a/  =4,  a;"  =  1 ;  y  =  0,  y  =  3. 

\y''  —  1xy  —  ^x  —  x^  =  0, 

I  y2_2xy  — 5x  — 2a;«  +  2  =  0.  _  

Am.  iK  =  1,  or  —  2,  y  =  1  ±  v/5,  or  —  2  ±x/—  4. 

\yx-  —  Zfx^  +  3y  V  —  Ay*x^  +  bifx  —  2y^  =  Q 
I  a;''  — 3yx  +  2/  =  0. 

Ans.  X  and  y  indeterminate. 

^S  —  2yx'  +  x'-\-x^  —  2yx+f  =  0, 

\  x"  —  2yx  +  1  =  0.  

Ans.  x  =  =i=  1;  or  dr^ —  3, 2/  =  =t  L 


506 
5. 


ELIMINATION    BETWEEN    TWO    EQUATIONS 


9. 


10. 


11. 


12. 


x^  —  2i/x^  ^x"  +  x"  —  2yx  +  1=0, 

a;2  _  22/^  +  1  =  0. 

Ans.  x  =  ^,  y=%. 

y""  —  2xy  -{■  x^  —  2y  —  1  z=  0, 
y^  —  2xt/  +  x'^  +  x  =  0. 

Tins.  x  =  —  1,  or  —  ^ ;  1/  =  0,  OT  —  |. 


2/2  _  2x1/  +  2x^  —  2y  +  4:  =  0, 
2/"  —  4xy  +  7a^  —  22/2  —  5  =  0. 


y^  —  ^xy  +  5x'  +  2cc  +  1  =  0, 
y  —  2a;  =  0. 


Ans.  Values  imaginary. 


cc  =  — l,y  =  — 2. 


3^2  +  CC2/  —  2a;2  +  3x  —  1  =  0, 

y'^  —  X^=  0. 

Ans.  35  =  J,  or  1,  or  I,  and  y  =  f ,  or  —  1,  or  —  J. 

(ya;-l)(x-2)(y_4)  =  0, 
(x  +  2)(2/x-a;).(a>  — 3)  =  0. 


J.ns. 


Values  of  x. 

Values  of  y. 

x  =  —2 

y  =  -J 

x^\ 

y=  +  l 

a;  =  3 

y=+i 

a;  =  0 

y  incompatible 

a;  =  2 

2,  =  1 

a;  incompatible 

y  incompatible 

a;  =  — 2 

y  =  4 

a;  =  0 

y  =  4 

a;  =  3 

y  =  4 

/  — 2ar2/  +  x''  +  2y=0, 
y2_  2xy  +  cc^  —  2y  —  1  =  0. 


},  a;  =  —  }±v'^. 


2^V  —  2/x*  —  S^x  —  2yx^  +  8yx  +  16y  +  2a;='  — 16  =  0, 

yX  +^  — 03— 1  =  0. 

"or 


Final  equation,  y^  +  y  —  2  =  0. 


OF    ANY    DEGREE. 


507 


13. 


y-^'  +  7  =  0,     (A) 
/_x^  +  3  =  0.     (B) 


Ans.  a;  =  2,  and  y  =  1. 


After  the  first  division,  multiply  (B)  by  (cc^  —  3)^,  we  will  get  for 
the  final  equation,  Qx"  —  Ux^  —  21a^  +  7Q  =  0.  The  only  rational 
value  in  this  equation  is  a;  =  2. 


14. 


y- 


15  =  0, 
7  =  0. 


Ans.  x  =  l,  and  y  =  2. 
Final  equation,  (x*  +  15)="  —  {x"  +  7)*  =  0. 
Required  the  values  and  the  final  equation  belonging  to 


15. 


I  ^6_a;s_2101  =  0, 
1  2/'  — X*  — 369  =  0. 


THE  END. 


-  THT 


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